How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$?

Question

The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$

First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n^2+n+1$ while my textbook has $a_n=n^2+2n+1$

Here is my work:

$ a_{n+1} - a_n = 2n+3 $

$pn=nd_1+d_2$ for all n

plug in and solve for $d_1$:

$(n+1)d_1+d_2-nd_1-d_2=2n+3$

$d_1=2n+3$

then solve for $d_2$:

$a_0=1=a_{n+1}-2n-3=(n+1)d_1+d_2-2n-3$

$d_2=1-n(2n+3)$

therefore

$p_n=nd_1+d_2 = 2n^2+n+4$ and $a_n=A+2n^2+n+4$

then solve for A:

$a_0=1=A +0+0+4$ so $A=-3$

therefore

$a_n=2n^2+n+1$

Answer 1

You got in trouble when you set up $p_n$. Since the forcing term is linear, we expect $p_n$ to be quadratic in $n$; $p_n=d_0n^2+d_1n+d_2$. Now we have

$$\begin{align*} 2n+3&=p_{n+1}-p_n\\ &=d_0(n+1)^2+d_1(n+1)+d_2-d_0n^2-d_1n-d_2\\ &=d_0(2n+1)+d_1\\ &=2d_0n+(d_0+d_1)\;. \end{align*}$$

This has to hold for all $n$, so we must have $2=2d_0$ and $3=d_0+d_1$. Solving this system is easy: $d_0=1$, so $d_1=2$, and $p_n=n^2+2n+d_2$. Now we want $p_0=a_0=1$, so $d_2=1$, and

$$p_n=n^2+2n+1\;.$$

Finally, $a_n=p_n+A=n^2+2n+1+A$, and setting $n=0$ shows that we must have $A=0$.

Alternatively, we could have left $d_2$ temporarily undetermined, writing $p_n=n^2+2n+d_2$, and observed that then

$$a_n=n^2+2n+d_2+A\;;$$

setting $n=0$ yields the conclusion that $d_2+A=1$ and hence $a_n=n^2+2n+1$. That is, it wasn’t actually necessary to solve separately for $d_2$ and $A$, since in the end we care only about their sum anyway.

Answer 2

If $ a_{i+1} - a_i = 2i+3 $ then $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=\sum_{i=0}^{n-1}(2i+3)$.

Observe that $\sum_{i=0}^{n-1}(a_{i+1} - a_i)=a_n-a_0$ and $\sum_{i=0}^{n-1}(2i+3)=n(n-1)+3n$.

Hence $a_n=1+n(n-1)+3n=(n+1)^2$.

Answer 3

Let $f(z) = \sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields $$\sum_{n=0}^\infty a_{n+1}z^n - \sum_{n=0}^\infty a_nz^n = \sum_{n=0}^\infty(2n+3)z^n $$ and hence $$z^{-1}(f(z)-a_0) - f(z) = 2\sum_{n=0}^\infty (n+1)z^n + \sum_{n=0}^\infty z^n, $$ so that $$f(z)(1-z) = 1 + \frac{2z}{(1-z)^2}+\frac z{1-z} $$ and \begin{align} f(z) &= \frac1{1-z}+\frac z{(1-z)^2} + \frac{2z}{(1-z)^3}\\ &= \frac{(1-z)^2+z(1-z)+2z}{(1-z)^3}\\ &= \frac{1 - 2z + z^2 + z -z^2 + 2z}{(1-z)^3}\\ &= \frac{1+z}{(1-z)^3}. \end{align} Expanding $f$ in a power series about $z=0$ yields $$f(z) = \sum_{n=0}^\infty(n+1)^2 z^n, $$ so that $a_n=(n+1)^2, n\geqslant0$.