n
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
i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2
i=1
n
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
i(2^i) = (k-1) * 2^(k+1) + 2 = ((k+1)-1 * 2^ ((k+1)+1 + 2
i=1