# Prove that for every positive integer n, ∑ k2^k = (n-1)2^(n+1) + 2?

Prove that for every positive integer n, ∑(from k =1 to n) k2^k = (n-1)2^(n+1) + 2 by INDUCTION.

My first step was to prove the basis step:

P(1) = 1*2^1 = (1-1)2^(1+1) + 2 = 2.

I have trouble with the inductive step:

First we assume the inductive hypothesis.

We'd have to prove ∑ k2^k + (k+1)2^(k+1) = k*2^(k+2)+2

And now I'm lost... please help!

### 3 Answers

- Anonymous9 years agoFavourite answer
Assume that it's true for n=j. Then:

∑ k (2^k) = (j-1) 2^(j+1) + 2

where the sum goes from n = 1 to j. If we add (j+1) * 2^(j+1) to both sides, we get the sigma going from n=1 to j+1, and:

∑ k (2^k) = (j-1) 2^(j+1) + 2 + (j+1) * 2^(j+1)

∑ k (2^k) = j 2^(j+1) - 2^(j+1) + 2 + j*2^(j+1) + 2^(j+1)

∑ k (2^k) = j 2^(j+1) + j*2^(j+1) + 2

∑ k (2^k) = 2j 2^(j+1) + 2

∑ k (2^k) = j 2^(j+1 + 1) + 2

∑ k (2^k) = (j+1 - 1) 2^(j+1 +1) + 2

So this proves that if it works for n=j, then it works for n=j+1 too.

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- Moja1981Lv 59 years ago
I'll be starting from the inductive step.

∑ k2^k + (k+1)2^(k+1)

: (k-1)2^(k+1) + 2 + (k+1)2^(k+1)

: k 2^(k+1) - 2^(k+1) + 2 + k 2^(k+1) + 2^(k+1)

: 2k 2^(k+1) + 2

: k 2^(k+1 + 1) + 2

: k 2^(k+2) + 2

Hope this helps!!!!!

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- Moise GunenLv 79 years ago
Step n is

∑ k2^k = (n-1)2^(n+1) + 2

Step n+1 is

∑ k2^k + (n+1)2^(n+1)= (n-1)2^(n+1) + 2 + (n+1)2^(n+1)

n+1

∑ k2^k = (n-1)2^(n+1) + 2 + (n+1)2^(n+1) = 2^(n+1)[n-1+n+1] +2 =2^(n+1)*2n +2 =

1

n2^(n+2) +2 QED

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