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
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
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