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a.(1/33)*sigma(n=1 to n=16) {(4*n^4)/((4*n^2)-1)} = ?
b.sigma(k=0 to k=n){(2^k)*(2nCk)} = ?
Dear student,
sigma(n=1 to n=16) {(4*n^4)/((4*n^2)-1)} =
4n^4/[(2n-1)(2n+1)] = 4n^4 -1+1/4n^2-1 = [4n^2 +1] + (1/2)[1/2n-1 - 1/2n+1]
sigma{[4n^2 +1]} = 4n(n+1)(2n+1)/6 + n put n= 16 ....... (1)
sigma{(1/2)[1/2n-1 - 1/2n+1]} = (1/2)[1+1/33] ....... (2)
add (1) and (2) gives solution
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