Deduce a formula (in terms of n) for the following sum: sum (2^i * i) where 1<=i<=n, n,i: natural numbers ( one can write this sum as: 1*2^1+ 2*2^2+ .. +n*2^n)

Let S = sum i* 2i (i=1,2..n) We multiply this by 2 and we obtain:2S = sum (i2i+1 ) = sum [(i+1-1) * 2i+1 ]= sum[ (i+1)2i+1 ] - sum 2i+1 i=1,2..n ()
sum (i+1)2i+1 = 222 + .. n
2n + (n+1)* 2n+1 = S + (n+1) * 2n+1 - 1*21 = S + (n+1)2n+1 -2
Note that the second sum is the sum of a geometric series of ratio 2, it is equal to 2n+2 - 4. Here is a quick proof:Let Z = sum 2i+1 = 22 + 23+ .. + 2n+1 . 2Z = sum 2i+2 = 23 + .. + 2n+1 + 2n+2 . So, 2Z-Z = Z = 2n+2 - 22
Going back to (
) we obtain that 2S =S + (n+1) * 2n+1 - 2 - (2n+2 - 4), which implies that S = (n+1) * 2n+1 - 2n+2 + 2


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Answered by Alexandru S. MAT tutor

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