I need help in simplifying the below double summation.
$\sum_{j=1}^i \sum_{k=i}^n (k-j+1) (2^{\max(j-2,0)})(2^{\max(n-k-1,0)})$
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Since this is not continuous sort of, I'd split it up into multiple sums, depending on the cases for you exponents. – Simply Beautiful Art Jan 02 '17 at 19:28
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I tried splitting up but I'm getting an incorrect solution. I think I'm doing something wrong while simplifying. That's why I ended up posting the question here. – Nilanshu96 Jan 02 '17 at 19:35
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$$\sum_{j=1}^i \sum_{k=i}^n (k-j+1) (2^{\max(j-2,0)})(2^{\max(n-k-1,0)}) = \sum_{j=1}^i 2^{\max{j-2}, 0}(1+\sum_{k=i}^{n-1}2^{n-k-1} =\sum_{j=1}^i 2^{\max{j-2}, 0} (1+\sum_{m=0}{n-1-i} 2^m) =\sum_{j=1}^i 2^{\max{j-2}, 0} ( 2^{n-1}-2^i) =(2+\sum_{k=0}^{i-2} 2^k)(2^{n-1}-2^i)).$$
Igor Rivin
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It can actually be more simplified to
S = 3*2^(n-1)-2^(n-i)-2^(i-1). But I'm not able to get this result. – Nilanshu96 Jan 04 '17 at 14:53