How to solve $V(n) = 2 \cdot V(n-1) + 2 \cdot n$

Question

How to solve $V(n) = 2 \cdot V(n-1) + 2\cdot n$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?

Answer 1

Try to expand and generalize the case (by induction): $$V(n) = 2\times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = \sum_{i=1}^{n}2^i(n-i+1)$$

Suppose $V(1) = 2$ in this case.

Answer 2

Let $V(m)=f(m)+am+b$

$$2n=V(n)-2V(n-1)=f(n)+an+b-2\{f(m-1)+a(n-1)+b\}$$ $$2n=f(n)-2f(n-1)-an+2a-b$$

Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0\le r\le n$

$\implies a=-2, b=2(-2)=?$