Find all solutions of the recurrence relation $a_n = 2a_{n−1} + 2n^2$

Question

The associated homogeneous recurrence relation is $a_n = 2a_{n−1}$. The characteristic equation is $r − 2 = 0$. Since our characteristic root is $r = 2$, we know by Theorem 3 that $a_n = α_{2}^n$ Note that $F(n) = 2n^2$ so we know by Theorem 6 that $s = 1$ and $1$ is not a root, the particular solution is of the form $n = c · 2n^2$ Plug into recurrence relation $c.2n^2 = 2.c(n-1)^2 + 2n^2$

is this correct? if then can we discuss further step to get value of c and solutions?

Answer 1

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

$$a_{n-1}=2a_{n-2}+2(n-1)^2$$

$$a_{n-2}=2a_{n-3}+2(n-2)^2$$

and so on, which tells you

$$a_n=2^2a_{n-2}+2^2(n-1)^2+2n^2$$

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

and so on down to

$$a_n=2^{n-1}a_1+\sum_{i=0}^{n-1}2^{i+1}(n-i)^2$$

Answer 2

Let $a_m=b_m+c_0+c_1m+c_2m^2+\cdots$

$$2n^2=b_n+c_0+c_1n+c_2n^2+\cdots-2(b_{n-1}+c_0+c_1(n-1)+c_2(n-1)^2+\cdots)$$

Clearly $c_r=0$ for $r\ge3$

Consequently $$2n^2=b_n-2b_{n-1}+c_0(1-2)+c_1(n-2(n-1))+c_2(n^2-2(n-1)^2))$$

Compare the coefficients of$x^0,x^1,x^2$ so that $$b_n=2b_{n-1}=\cdots=2^rb_{n-r},r\le n-1$$

Answer 3

..a_n= a_n(h) + a_n(particular soln)

 =c.(2)^n + { p2(n^2) + p1.n + p0 } (1)^n

 = c.(2)^n + p2(n^2) + p1.n + p0 

Then we plug in the particular soln in the recurrence relation.. and get

(P2+2)n^2 +(p1-4p2)n + (2p2-2p1+p0)=0 And so for the eqn to be true for all n P2+2 =0 P1-4p2 =0 2p2-2p1+p0=0

Solving we get p0=-12, p1=-8,p2=-2

So all solns are of the form ..a_n =c.(2)^n -2(n^2) -8n -12

This is the exact soution.