Find the sum of: 4 1 − 1 2 1 + 4 2 − 1 2 2 + 4 4 − 1 2 4 + 4 8 − 1 2 8 + …
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Is it possible to have 2 variables in one sum? k = 0 ∑ ∞ ( 2 1 ) ( 1 + 2 k ) 2 n You have 2 variables; k and n
I got wrong for this question but I managed to write up to this (after that I was lost)--[:(]
n = 2 x ∑ ∞ ( 4 n − 1 2 n )
:))
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It's not a sum over two variables: k is the only bound variable in that sum. If n = 3 , the sum is k = 0 ∑ ∞ ( 2 1 ) ( 1 + 2 k ) 2 3 or if n = 1 2 , the sum is k = 0 ∑ ∞ ( 2 1 ) ( 1 + 2 k ) 2 1 2 . This is what I was saying at the beginning with the first line simply being the n th term in the original sum.
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The general term of the sum, indexed by n = 0 , 1 , 2 , … is 2 2 ⋅ 2 n − 1 2 2 n = 1 − 2 − 2 ⋅ 2 n 2 − 2 n = k = 0 ∑ ∞ ( 2 1 ) ( 1 + 2 k ) 2 n so the sum in the problem is n = 0 ∑ ∞ k = 0 ∑ ∞ ( 2 1 ) ( 1 + 2 k ) 2 n = m = 1 ∑ ∞ ( 2 1 ) m where the equality is by rearranging (all the terms are positive, so that's no problem), using the fact that every positive integer is written uniquely as the product of a non-negative power of 2 and a positive odd number.
Therefore the sum is m = 1 ∑ ∞ ( 2 1 ) m = 1 − 2 1 2 1 = 1