n

$\displaystyle \sum$ i(2^i) = (n-1) * 2^(n+1) + 2

i=1

So far if have the base case 1(2^1) = (1-1) * (1+1) + 2

Induction part is giving me problems:

k

$\displaystyle \sum$ i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2

i=1

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- Aug 31st 2010, 07:50 PMzaj091000Proof by Induction
n

$\displaystyle \sum$ i(2^i) = (n-1) * 2^(n+1) + 2

i=1

So far if have the base case 1(2^1) = (1-1) * (1+1) + 2

Induction part is giving me problems:

k

$\displaystyle \sum$ i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2

i=1 - Aug 31st 2010, 08:49 PMCaptainBlack
Assume that what was to be proved holds for n=k, then:

$\displaystyle \displaystyle \sum_{i=1}^k i 2^i=(k-1)2^{k+1}+2$

Add the next term from the left to both sides:

$\displaystyle \displaystyle (k+1)2^{k+1}+\sum_{i=1}^k i 2^i =\sum_{i=1}^{k+1} i 2^i =(k-1)2^{k+1}+2 + (k+1)2^{k+1}$

Now simplify the right hand side.

CB