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
Last edited by zaj091000; August 31st 2010 at 09:14 PM.
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Originally Posted by zaj091000 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 Assume that what was to be proved holds for n=k, then: Add the next term from the left to both sides: Now simplify the right hand side. CB
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