Recurrence Relation Calculator (General Term with Steps)

Enter the first term a₁, the coefficient p and the constant q of the recurrence relation a(n+1) = p·a(n) + q. The formula below is linked to the input fields, so you can also edit the numbers right inside it.

First term a₁

Coefficient p (multiplies aₙ)

Constant q (added)

Term number n

Enter numbers only. Decimals, negative numbers and fractions such as 1/2 are OK. If "Term number n" is left blank, it is treated as 10.

Result and graph

Enter the first term and the coefficients in the fields on the left and press "Calculate". The result and a graph will appear here.

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What you can do on this page
What is this calculation used for?
How to Use
Formulas and graphs
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Finds the general term of the first-order recurrence relation $a_{n+1} = p a_n + q$ (with first term $a_1$) on the spot, with steps that use the characteristic equation to turn it into a geometric sequence
Answers are exact fractions in lowest terms, such as $\dfrac{381}{64}$. The special cases $p = 1$ (arithmetic sequence) and $q = 0$ (geometric sequence) are handled automatically
Also shows the value of the term at the position $n$ you choose (the $n$th term), plus a table and graph of terms 1 to 10. If the sequence converges, you also get the value it approaches (the limit, which is the equilibrium value)
The first term and the coefficients can be decimals, negative numbers or fractions such as 1/2
The calculator shows the recurrence relation just as you would write it by hand, and you can edit the numbers right inside it
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

This page handles the first-order recurrence relation $a_{n+1} = p a_n + q$ ($p$ and $q$ are constants). It does not handle forms that contain $n$, such as $a_{n+1} = a_n + n$, or second-order recurrence relations that involve $a_{n+2}$.

What is this calculation used for?

Track how a loan balance goes down each month (money and life planning)

For a loan that charges interest every month and is paid down by a fixed amount every month, the balance moves by exactly the first-order recurrence relation "next month's balance = (1 + monthly rate) × this month's balance − monthly payment".
For example, with a monthly rate of 0.5% and a monthly payment of $1,500, the equilibrium value is 1500 ÷ 0.005 = 300,000 dollars. This is the loan amount at which interest and payments balance exactly and the balance stops moving. If the balance is below it, every payment makes it go down for sure (this balance idea is the basis of repayment planning).

Estimate where the level of a daily medicine settles (medicine and pharmacy)

The body breaks down and removes a medicine at a steady rate over time. In a model where "50% of the medicine is still in the body after one day, and you take 200 mg every day", the amount in the body follows the recurrence relation $a_{n+1} = 0.5a_n + 200$ and converges to the equilibrium value 200 ÷ (1 − 0.5) = 400 mg.
This formula shows that even if you keep taking it every day, it does not build up forever; it settles at a steady level. In real drug design, doses and intervals are set with pharmacokinetic models that make this kind of calculation more precise.

Where a subscription's member count levels off (business)

For a service where "90% of members stay each month and 1,000 new people sign up", the member count follows the recurrence relation $a_{n+1} = 0.9a_n + 1000$ and converges to the equilibrium value 1000 ÷ (1 − 0.9) = 10,000 people.
So if sign-ups and the churn rate stay the same, membership levels off at about 10,000. This is a basic business-planning model for estimating the ceiling on growth and how much higher that ceiling goes if you lower the churn rate (a rough guide that assumes the retention rate and sign-ups stay constant).

How a room warms up with the heat on (home and physics)

When you heat a cold room, the gap between the room temperature and the thermostat setting shrinks by a steady percentage every minute. For example, if "every minute, the gap between the room temperature and the setting of 77°F shrinks by 10%", the room temperature follows the recurrence relation $a_{n+1} = 0.9a_n + 7.7$ and converges to the equilibrium value 7.7 ÷ 0.1 = 77°F, the thermostat setting.
"It warms up quickly at first, then more slowly as it gets close to the setting" is exactly what converging looks like for this recurrence relation (in physics, it is a minute-by-minute version of Newton's law of cooling).

Write the growth of a savings plan as a formula (money and investing)

The balance of a savings plan where "interest is added at the end of each month, and you also deposit a fixed amount every month" follows the same kind of recurrence relation: "next month's balance = (1 + monthly rate) × this month's balance + deposit". Here, though, the multiplier is greater than 1, so it does not converge and the balance keeps growing (an example that diverges).
Finding the general term lets you calculate "the balance after n months" in one step. It is the same as the future value of an annuity formula used in finance. Estimates of how much you will save are made with the general term of this recurrence relation.

Formulas and graphs

First-order recurrence relation

Graph

Standard notation (the usual math form)

$a_{n+1}$ $=$ $p$ $\times$ $a_n$ $+$ $q$

In words (symbols replaced with words)

④ $a_{n+1}$: next term $=$ ② $p$: multiplier $\times$ ① $a_n$: current term $+$ ③ $q$: number added

The formula in words

① Take the $a_n$: current term , multiply it by the

② $p$: multiplier , add the

③ $q$: number added ,

④ and you get the $a_{n+1}$: next term

Quick example

With first term $a_1 = 1$ and the recurrence relation $a_{n+1} = 2a_n + 3$ (double the current term and add 3), the sequence starts like this

next term $=$ 2 $\times$ current term $+$ 3

$a_2 = 2 \times 1 + 3 = 5$

$a_3 = 2 \times 5 + 3 = 13$

$a_4 = 2 \times 13 + 3 = 29$

Key idea

A recurrence relation is a rule that makes the next term from the current term. With the first term $a_1$ and the rule, the whole sequence $a_2$, $a_3$, … follows in order. But to find term 100 this way, you would have to calculate 99 times. That is why we want the general term, a formula that gives the $n$th term directly from $n$. The tool for finding it is the characteristic equation, explained next.

The characteristic equation and the equilibrium value

Standard notation (the usual math form)

$c$ $=$ $q$ $\div$ $1 - p$

In words (symbols replaced with words)

③ $c$: equilibrium value $=$ ① $q$: number added $\div$ ② $1 - p$: 1 minus the multiplier $p$

The formula in words

① Replace both $a_{n+1}$ and $a_n$ in the recurrence relation with $x$ and solve the equation $x = px + q$ (the characteristic equation). Divide the $q$: number added by

② $1 - p$: 1 minus the multiplier $p$ ,

③ and you get the $c$: equilibrium value

Quick example

The characteristic equation of the recurrence relation $a_{n+1} = 2a_n + 3$ is $x = 2x + 3$. Solving it gives

$c$: equilibrium value $=$ number added (3) $\div$ $1 - 2$

$x = 2x + 3$

$x - 2x = 3$

$-x = 3$

$x = -3$

$c = \dfrac{3}{1-2} = -3$

Key idea

The solution $c$ of the characteristic equation $x = px + q$ is a special value: if a term ever equals it, the next term is exactly the same value (indeed $c = pc + q$, so after $c$ the sequence stays at $c$). For this reason, the equilibrium value is also called the fixed point of the sequence. You can solve the equation by moving terms, or just use the formula $c = \dfrac{q}{1-p}$. Since the denominator is $1 - p$, the characteristic equation does not work only when $p = 1$ (that case is an arithmetic sequence, covered in a separate formula card below).

Rewriting with the equilibrium value (making a geometric sequence)

Standard notation (the usual math form)

$a_{n+1} - c$ $=$ $p$ $\times$ $(a_n - c)$

In words (symbols replaced with words)

③ $a_{n+1} - c$: gap between the next term and the equilibrium value $=$ ② $p$: multiplier $\times$ ① $a_n - c$: gap between the current term and the equilibrium value

The formula in words

① Multiply the $a_n - c$: gap between the current term and the equilibrium value by the

② $p$: multiplier ,

③ and you get the $a_{n+1} - c$: gap between the next term and the equilibrium value (looking only at the gaps, they form a geometric sequence with common ratio $p$)

Quick example

Rewriting the recurrence relation $a_{n+1} = 2a_n + 3$ (equilibrium value $c = -3$) gives

next term $+\,3$ $=$ 2 $\times$ current term $+\,3$

$a_{n+1} + 3 = 2\left(a_n + 3\right)$

$a_1 + 3 = 4,\quad a_2 + 3 = 8,\quad a_3 + 3 = 16$

Key idea

Subtract the characteristic equation $c = pc + q$ from the recurrence relation $a_{n+1} = p a_n + q$, side by side (left side minus left side, right side minus right side). The number added, $q$, cancels out, and only $a_{n+1} - c = p(a_n - c)$ is left. So the sequence of gaps $\{a_n - c\}$, each term minus the equilibrium value $c$, is a geometric sequence that just multiplies by $p$ each time. The goal of this rewrite is to turn a hard-to-handle recurrence relation that mixes in an addition into a familiar geometric sequence. In the example, $c = -3$, so the gap is $a_n - (-3) = a_n + 3$, and it goes $4,\ 8,\ 16,\ \dots$, doubling each time as expected.

General term (characteristic equation case)

Graph

Standard notation (the usual math form)

$a_n$ $=$ $(a_1 - c)$ $\times$ $p^{n-1}$ $+$ $c$

In words (symbols replaced with words)

④ $a_n$: $n$th term (general term) $=$ ① $a_1 - c$: gap between the first term and the equilibrium value $\times$ ② $p^{n-1}$: the multiplier $p$ multiplied $n-1$ times $+$ ③ $c$: equilibrium value

The formula in words

① Multiply the $a_1 - c$: gap between the first term and the equilibrium value by

② $p^{n-1}$: the multiplier $p$ multiplied $n-1$ times , add back the

③ $c$: equilibrium value ,

④ and you get the $a_n$: $n$th term (general term)

Quick example

With first term $a_1 = 1$ and the recurrence relation $a_{n+1} = 2a_n + 3$ (equilibrium value $c = -3$), the general term is

$n$th term $=$ gap $1 - (-3) = 4$ $\times$ $2^{n-1}$ $+$ equilibrium value $-3$

$a_n = \left(1 - (-3)\right) \times 2^{n-1} + (-3) = 4 \cdot 2^{n-1} - 3$

$a_1 = 4 \times 1 - 3 = 1,\quad a_2 = 4 \times 2 - 3 = 5,\quad a_3 = 4 \times 4 - 3 = 13$

Key idea

The geometric sequence $\{a_n - c\}$ from the rewrite has first term $a_1 - c$ and common ratio $p$, so by the general-term formula for geometric sequences, $a_n - c = (a_1 - c)p^{n-1}$. Finally, add back the $c$ you subtracted (move it to the other side), and you get this general term. When $|p| < 1$ ($p$ is between $-1$ and $1$), $p^{n-1}$ gets closer to 0 as $n$ grows, so $a_n$ gets closer and closer to the equilibrium value $c$ (it converges). When the sequence converges, this calculator shows $c$ as the limit.

When $p = 1$ (arithmetic sequence)

Standard notation (the usual math form)

$a_n$ $=$ $a_1$ $+$ $(n-1)q$

In words (symbols replaced with words)

③ $a_n$: $n$th term (general term) $=$ ① $a_1$: first term $+$ ② $(n-1)q$: the common difference $q$ added $n-1$ times

The formula in words

① When $p = 1$, the recurrence relation is $a_{n+1} = a_n + q$ (it just adds $q$ each time). So add to the $a_1$: first term the

② $(n-1)q$: the common difference $q$ added $n-1$ times ,

③ and you get the $a_n$: $n$th term (general term)

Quick example

With first term $a_1 = 5$ and the recurrence relation $a_{n+1} = a_n + 3$ (an arithmetic sequence with common difference 3), the general term and term 4 are

$n$th term $=$ first term (5) $+$ $3$ added $n-1$ times

$a_n = 5 + (n-1) \times 3 = 3n + 2$

$a_4 = 3 \times 4 + 2 = 14$

Key idea

When $p = 1$, the denominator $1 - p$ in the characteristic equation becomes 0, so the equilibrium value cannot be used. Instead, the recurrence relation just adds the same number $q$ each time, so the sequence is simply an arithmetic sequence with common difference $q$. From term 1 to term $n$, you add $q$ as many times as there are gaps between terms: $n-1$ times (from term 1 to term 4 is 3 times; counting it as $n$ times is a common mistake).

When $q = 0$ (geometric sequence)

Standard notation (the usual math form)

$a_n$ $=$ $a_1$ $\times$ $p^{n-1}$

In words (symbols replaced with words)

③ $a_n$: $n$th term (general term) $=$ ① $a_1$: first term $\times$ ② $p^{n-1}$: the common ratio $p$ multiplied $n-1$ times

The formula in words

① When $q = 0$, the recurrence relation is $a_{n+1} = p a_n$ (it just multiplies by $p$ each time). So multiply the $a_1$: first term by

② $p^{n-1}$: the common ratio $p$ multiplied $n-1$ times ,

③ and you get the $a_n$: $n$th term (general term)

Quick example

With first term $a_1 = 3$ and the recurrence relation $a_{n+1} = 2a_n$ (a geometric sequence with common ratio 2), the general term and term 4 are

$n$th term $=$ first term (3) $\times$ $2^{n-1}$

$a_n = 3 \times 2^{n-1}$

$a_4 = 3 \times 2^{3} = 3 \times 8 = 24$

Key idea

When $q = 0$, the recurrence relation just multiplies by the same number $p$ each time, so the sequence is simply a geometric sequence with common ratio $p$. For the same reason as in the arithmetic case, you multiply $n-1$ times. In fact, if you set $q = 0$ in the general term for the characteristic equation case, the equilibrium value is $c = 0$, and $a_n = (a_1 - 0)p^{n-1} + 0 = a_1 p^{n-1}$. So this formula is a special case of that one.

For the first-order recurrence relation $a_{n+1} = p a_n + q$, use the solution of the characteristic equation $x = px + q$ (the equilibrium value $c = \dfrac{q}{1-p}$) to rewrite it as $a_{n+1} - c = p(a_n - c)$. The sequence of gaps $\{a_n - c\}$ is then geometric with common ratio $p$, which gives the general term $a_n = (a_1 - c)p^{n-1} + c$. If $p = 1$, it is an arithmetic sequence; if $q = 0$, it is a geometric sequence.

Symbols and terms

Symbols

$a_n$	a sub n	The $n$th term of the sequence, counting from the start. The letter $a$ is often used for sequences (said to be because it is the first letter of the alphabet). The small subscript $n$ at the lower right tells which position it is.
$a_{n+1}$	a sub n plus one	The term right after the $n$th term, that is, term $n+1$. In a recurrence relation it plays the role of "the next term after the current term $a_n$".
$a_1$	a sub one	The first term of the sequence. A recurrence relation is a rule that makes the next term from the current one, so the sequence is not fixed until its starting point, the first term, is set.
$n$	en	The position number of a term (how far from the start). By custom it is $n$, for "number". It takes the values 1, 2, 3, … (natural numbers).
$p$	pee	The constant that multiplies the current term $a_n$. It plays the role of the common ratio of a geometric sequence. If it is between $-1$ and $1$, the sequence converges to the equilibrium value.
$q$	cue	The constant added after multiplying. It plays the role of the common difference of an arithmetic sequence. By custom, $q$ is used as the letter after $p$.
$c$	cee	The solution of the characteristic equation $x = px + q$ (the equilibrium value). The letter $c$ comes from "constant". It is a special value: once a term equals it, the next term stays the same. It is also where a converging sequence ends up.
$\{a_n\}$	the sequence a sub n	A way to write the whole sequence made of the terms $a_n$. The curly braces say "all of $a_1, a_2, a_3, \dots$ taken together".
$p^{n-1}$	p to the power of n minus one	$p$ multiplied by itself $n-1$ times (a power). The small number at the upper right (the exponent) tells how many times to multiply. From term 1 to term $n$ you multiply by the common ratio $n-1$ times, which is why this form appears in the general term.
$\displaystyle\lim_{n\to\infty} a_n$	the limit of a sub n as n approaches infinity	The value $a_n$ approaches as $n$ grows without bound (the limit). $\lim$ is the first three letters of "limit". This calculator shows the limit when the sequence converges.

Terms

sequence	Numbers lined up in order. Each number is called a term, and terms are numbered, as in $a_1, a_2, a_3, \dots$.
term	Each single number in a sequence. The one in position $n$ from the start is called the $n$th term.
first term	The first term $a_1$ of a sequence. It is the starting point when a recurrence relation defines a sequence.
recurrence relation (recursive formula)	A formula for the rule that makes the next term from the current term. For example, $a_{n+1} = 2a_n + 3$ is the rule "the next term is the current term doubled, plus 3". Together with the first term, it fixes the whole sequence.
first-order recurrence relation	A recurrence relation written with only two neighboring terms ($a_n$ and $a_{n+1}$). The $a_{n+1} = p a_n + q$ on this page is the typical form. One with three terms $a_n,\ a_{n+1},\ a_{n+2}$ is called a second-order recurrence relation and is solved differently.
general term (explicit formula)	The $n$th term written directly as a formula in $n$. For example, once you have $a_n = 4 \cdot 2^{n-1} - 3$, even term 100 takes one step: just put in $n = 100$.
arithmetic sequence	A sequence where the difference between neighboring terms is always the same (it adds the same number each time). $5, 8, 11, 14, \dots$ is an arithmetic sequence with common difference 3. It is the case $p = 1$ of the recurrence relation.
common difference	The fixed number an arithmetic sequence adds each time. In the recurrence relation on this page, $q$ plays this role when $p = 1$.
geometric sequence	A sequence where the ratio between neighboring terms is always the same (it multiplies by the same number each time). $3, 6, 12, 24, \dots$ is a geometric sequence with common ratio 2. It is the case $q = 0$ of the recurrence relation.
common ratio	The fixed number a geometric sequence multiplies by each time. In the recurrence relation on this page, $p$ plays this role.
characteristic equation	The equation $x = px + q$ you get by replacing both $a_{n+1}$ and $a_n$ in the recurrence relation with the same letter $x$ (also called the fixed-point equation). Its solution (the equilibrium value) lets you rewrite the recurrence relation as a geometric sequence.
equilibrium value	The solution $c = \dfrac{q}{1-p}$ of the characteristic equation. It is the point where the sequence stops moving: once a term equals it, the next term stays the same. When the sequence converges, it is also where the sequence ends up (the limit).
fixed point	A value that does not change when you apply the rule. For a recurrence relation, the equilibrium value $c$ is the fixed point, since $c = pc + q$. In college math, recurrence relations and repeated functions are studied from this point of view.
converge	As $n$ grows, the terms get closer and closer to one fixed value. This recurrence relation converges to the equilibrium value $c$ when $\|p\| < 1$ ($p$ is between $-1$ and $1$).
limit	The value a converging sequence approaches. It is written $\displaystyle\lim_{n\to\infty} a_n$. Limits are studied in depth in calculus, but thinking of it as "the destination the terms keep getting closer to" is enough here.
diverge	To not converge. This covers terms that grow without bound (or fall without bound) and terms that swing back and forth without settling on one value. This recurrence relation diverges when $\|p\| > 1$, unless it starts exactly at the equilibrium value.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back to review these topics is the fastest way forward.

What a sequence is (Algebra 1)	Knowing that a sequence is numbers lined up in order, and $a_n$ is "the $n$th number from the start (the $n$th term)" Being able to read expressions in subscripts (for example, $a_{n+1}$ is "the term after the $n$th term")
Arithmetic and geometric sequences (Algebra 1 and 2)	Being able to use the general term of an arithmetic sequence (add the same number each time), $a_n = a_1 + (n-1)d$ Being able to use the general term of a geometric sequence (multiply by the same number each time), $a_n = a_1 r^{n-1}$ Knowing why you add or multiply $n-1$ times, not $n$ times (it is the number of gaps between terms)
Solving linear equations (Grades 7–8)	Being able to solve an equation like $x = 2x + 3$ by moving terms to one side Knowing that you may subtract the same thing from both sides of an equation (used when subtracting the characteristic equation, side by side)
Powers and exponents (Grade 6 to Algebra 1)	Knowing that the small number at the upper right (the exponent) tells how many times to multiply, as in $2^{3} = 2 \times 2 \times 2 = 8$ Knowing the basic exponent rules, such as $2^{0} = 1$
Working with fractions (Grades 5–7)	Being able to multiply, divide and simplify fractions Being comfortable leaving answers as fractions (not being put off by an answer like $\dfrac{381}{64}$)

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the nth term with the general term (characteristic equation case)

First term a1	1
Coefficient p	2
Constant q	3
Term number n	10
Equilibrium value c = q/(1−p)	=B3/(1-B2)
Gap between first term and c, a1−c	=B1-B5
nth term (a1−c)×p^(n−1)+c	=B6*B2^(B4-1)+B5

Table that follows the recurrence relation step by step (terms 1–10)

Coefficient p	2
Constant q	3
Term 1, a1	1
Term 2	=B1*B3+B2
Term 3	=B1*B4+B2
Term 4	=B1*B5+B2
Term 5	=B1*B6+B2
Term 6	=B1*B7+B2
Term 7	=B1*B8+B2
Term 8	=B1*B9+B2
Term 9	=B1*B10+B2
Term 10	=B1*B11+B2

Table to find the nth term when p = 1 (arithmetic sequence)

First term a1	5
Common difference q	3
Term number n	4
nth term a1+(n−1)q	=B1+(B3-1)*B2

Table to find the nth term when q = 0 (geometric sequence)

First term a1	3
Common ratio p	2
Term number n	4
nth term a1×p^(n−1)	=B1*B2^(B3-1)

After pasting, the upper rows (first term and coefficients) are the inputs, and the lower rows show the calculated results. "^" is a power and "*" is multiplication.
The first table uses the general term to find term 10 of the sequence with first term 1 and recurrence relation a(n+1) = 2a(n) + 3. The equilibrium value c is −3, and the answer is 2045. The formulas in this table do not work when p = 1 (1−p becomes 0, and you would divide by 0). In that case, use the third table.
The second table calculates the same sequence step by step, exactly as the recurrence relation says, so you can check that term 10 is 2045, the same as in the first table.
The third table is an example of the arithmetic case (first term 5, common difference 3, term 4 = 14), and the fourth table is an example of the geometric case (first term 3, common ratio 2, term 4 = 24).

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the nth term with the general term (characteristic equation case)

First term a1	1
Coefficient p	2
Constant q	3
Term number n	10
Equilibrium value c = q/(1−p)	=B3/(1-B2)
Gap between first term and c, a1−c	=B1-B5
nth term (a1−c)×p^(n−1)+c	=B6*B2^(B4-1)+B5

Table that follows the recurrence relation step by step (terms 1–10)

Coefficient p	2
Constant q	3
Term 1, a1	1
Term 2	=B1*B3+B2
Term 3	=B1*B4+B2
Term 4	=B1*B5+B2
Term 5	=B1*B6+B2
Term 6	=B1*B7+B2
Term 7	=B1*B8+B2
Term 8	=B1*B9+B2
Term 9	=B1*B10+B2
Term 10	=B1*B11+B2

Table to find the nth term when p = 1 (arithmetic sequence)

First term a1	5
Common difference q	3
Term number n	4
nth term a1+(n−1)q	=B1+(B3-1)*B2

Table to find the nth term when q = 0 (geometric sequence)

First term a1	3
Common ratio p	2
Term number n	4
nth term a1×p^(n−1)	=B1*B2^(B3-1)

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the first term and coefficients with your own numbers.

How to calculate it in Python

from fractions import Fraction

first_term = Fraction(1)    # first term a1 (fractions work too, e.g. Fraction(1, 2))
coefficient = Fraction(2)   # coefficient p (multiplies a_n)
constant = Fraction(3)      # constant q (added)
term_number = 10            # term number n to find

# Calculate terms 1 to 10 in order, following the recurrence relation
terms = [first_term]
for _ in range(9):
    terms.append(coefficient * terms[-1] + constant)
print("Terms 1-10:", ", ".join(str(t) for t in terms))

# nth term (apply the recurrence relation n-1 times)
nth_term = first_term
for _ in range(term_number - 1):
    nth_term = coefficient * nth_term + constant
print(f"Term {term_number}: {nth_term} (as a decimal: {float(nth_term)})")

# If p is not 1, the equilibrium value c = q/(1-p) also gives the general term
if coefficient != 1:
    fixed_point = constant / (1 - coefficient)
    start_diff = first_term - fixed_point
    print(f"Equilibrium value c: {fixed_point}")
    print(f"General term: a_n = ({start_diff})*({coefficient})^(n-1) + ({fixed_point})")

With the fractions module from the standard library, you can calculate with exact fractions and no decimal rounding error. This example is the sequence with first term 1 and recurrence relation a(n+1) = 2a(n) + 3. Running it prints terms 1 to 10 (1, 5, 13, …, 2045), term 10 = 2045, the equilibrium value −3, and the general term a_n = (4)*(2)^(n-1) + (-3) (simplified, 4×2^(n−1) − 3). Change the first term and coefficients and run it.

How to write it in LaTeX and other math languages (copy and paste)

First-order recurrence relation

aₙ₊₁ = p·aₙ + q

a_{n+1} = p a_n + q

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>a</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msub>
    <mo>=</mo>
    <mi>p</mi>
    <msub><mi>a</mi><mi>n</mi></msub>
    <mo>+</mo>
    <mi>q</mi>
  </mrow>
</math>

a_(n+1) = p a_n + q

RSolve[{a[n + 1] == p a[n] + q, a[1] == a1}, a[n], n]

rsolve({a(n+1) = p*a(n) + q, a(1) = a1}, a(n));

a(n+1) = p*a(n) + q

a_(n+1) = p a_n + q

The characteristic equation and the equilibrium value

c = q/(1 − p)

c = \dfrac{q}{1-p}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>c</mi>
    <mo>=</mo>
    <mfrac>
      <mi>q</mi>
      <mrow><mn>1</mn><mo>&#x2212;</mo><mi>p</mi></mrow>
    </mfrac>
  </mrow>
</math>

c = q/(1-p)

c = q/(1 - p)

c := q/(1 - p);

c = q/(1 - p);

c = q/(1 − p)

Rewriting with the equilibrium value (making a geometric sequence)

aₙ₊₁ − c = p(aₙ − c)

a_{n+1} - c = p(a_n - c)

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>a</mi><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></msub>
    <mo>&#x2212;</mo><mi>c</mi>
    <mo>=</mo>
    <mi>p</mi>
    <mo>(</mo>
    <msub><mi>a</mi><mi>n</mi></msub>
    <mo>&#x2212;</mo><mi>c</mi>
    <mo>)</mo>
  </mrow>
</math>

a_(n+1) - c = p(a_n - c)

a[n + 1] - c == p (a[n] - c)

a(n+1) - c = p*(a(n) - c);

a(n+1) - c = p*(a(n) - c)

a_(n+1) − c = p(a_n − c)

General term (characteristic equation case)

aₙ = (a₁ − c)·pⁿ⁻¹ + c

a_n = (a_1 - c)p^{n-1} + c

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>a</mi><mi>n</mi></msub>
    <mo>=</mo>
    <mo>(</mo>
    <msub><mi>a</mi><mn>1</mn></msub>
    <mo>&#x2212;</mo><mi>c</mi>
    <mo>)</mo>
    <msup><mi>p</mi><mrow><mi>n</mi><mo>&#x2212;</mo><mn>1</mn></mrow></msup>
    <mo>+</mo>
    <mi>c</mi>
  </mrow>
</math>

a_n = (a_1 - c)p^(n-1) + c

a[n_] := (a1 - c) p^(n - 1) + c

a := n -> (a1 - c)*p^(n-1) + c;

a_n = (a1 - c)*p^(n-1) + c;

a_n = (a_1 − c)p^(n−1) + c

When $p = 1$ (arithmetic sequence)

aₙ = a₁ + (n − 1)q

a_n = a_1 + (n-1)q

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>a</mi><mi>n</mi></msub>
    <mo>=</mo>
    <msub><mi>a</mi><mn>1</mn></msub>
    <mo>+</mo>
    <mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mn>1</mn><mo>)</mo>
    <mi>q</mi>
  </mrow>
</math>

a_n = a_1 + (n-1)q

a[n_] := a1 + (n - 1) q

a := n -> a1 + (n-1)*q;

a_n = a1 + (n-1)*q;

a_n = a_1 + (n − 1)q

When $q = 0$ (geometric sequence)

aₙ = a₁·pⁿ⁻¹

a_n = a_1 p^{n-1}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>a</mi><mi>n</mi></msub>
    <mo>=</mo>
    <msub><mi>a</mi><mn>1</mn></msub>
    <msup><mi>p</mi><mrow><mi>n</mi><mo>&#x2212;</mo><mn>1</mn></mrow></msup>
  </mrow>
</math>

a_n = a_1 p^(n-1)

a[n_] := a1 p^(n - 1)

a := n -> a1*p^(n-1);

a_n = a1*p^(n-1);

a_n = a_1 p^(n−1)

How to have ChatGPT do the calculation

You are an assistant for math calculations (sequences and recurrence relations). Do the following calculations by actually running Python code, and base your answer only on the numbers from the output (do not answer from mental math or guesses).

For the sequence defined by the first term a_1 = 1 and the recurrence relation a_(n+1) = 2·a_n + 3, show each of the following.
1. The solution of the characteristic equation x = 2x + 3 (the equilibrium value c)
2. The general term a_n (with the steps that use c to rewrite it as a geometric sequence)
3. The values of terms 1 to 10, and the value of term 10
4. Does this sequence converge? (If it does, give the limit too.)

In Python, use the fractions module from the standard library to calculate exactly, and show the formulas you used and the numbers from the output.

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\(a_n\)	a sub n	The \(n\)th term of the sequence, counting from the start. The letter \(a\) is often used for sequences (said to be because it is the first letter of the alphabet). The small subscript \(n\) at the lower right tells which position it is.
\(a_{n+1}\)	a sub n plus one	The term right after the \(n\)th term, that is, term \(n+1\). In a recurrence relation it plays the role of "the next term after the current term \(a_n\)".
\(a_1\)	a sub one	The first term of the sequence. A recurrence relation is a rule that makes the next term from the current one, so the sequence is not fixed until its starting point, the first term, is set.
\(n\)	en	The position number of a term (how far from the start). By custom it is \(n\), for "number". It takes the values 1, 2, 3, … (natural numbers).
\(p\)	pee	The constant that multiplies the current term \(a_n\). It plays the role of the common ratio of a geometric sequence. If it is between \(-1\) and \(1\), the sequence converges to the equilibrium value.
\(q\)	cue	The constant added after multiplying. It plays the role of the common difference of an arithmetic sequence. By custom, \(q\) is used as the letter after \(p\).
\(c\)	cee	The solution of the characteristic equation \(x = px + q\) (the equilibrium value). The letter \(c\) comes from "constant". It is a special value: once a term equals it, the next term stays the same. It is also where a converging sequence ends up.
\(\{a_n\}\)	the sequence a sub n	A way to write the whole sequence made of the terms \(a_n\). The curly braces say "all of \(a_1, a_2, a_3, \dots\) taken together".
\(p^{n-1}\)	p to the power of n minus one	\(p\) multiplied by itself \(n-1\) times (a power). The small number at the upper right (the exponent) tells how many times to multiply. From term 1 to term \(n\) you multiply by the common ratio \(n-1\) times, which is why this form appears in the general term.
\(\displaystyle\lim_{n\to\infty} a_n\)	the limit of a sub n as n approaches infinity	The value \(a_n\) approaches as \(n\) grows without bound (the limit). \(\lim\) is the first three letters of "limit". This calculator shows the limit when the sequence converges.

Recurrence Relation Calculator (General Term with Steps)

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Arithmetic Sequence Calculator (nth Term and Sum)

Finds the nth term and the sum of the first n terms from the first term, common difference and number of terms (a sequence that adds the same number each time, the case p = 1 of this recurrence)

Geometric Sequence Calculator (nth Term and Sum)

Finds the nth term and the sum of the first n terms from the first term, common ratio and number of terms (a sequence that multiplies by the same number each time, the case q = 0 of this recurrence)

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