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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?

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  • $\begingroup$ Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done... $\endgroup$
    – Did
    Jan 20, 2019 at 11:03
  • $\begingroup$ @user623855 2n ≠ 2^n. $\endgroup$
    – Did
    Jan 20, 2019 at 11:04

2 Answers 2

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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.

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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 $f(n)=2f(n-1)=2^rf(n-r),0\le r\le n$

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

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