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)} = ?

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Pramod J AskiitiansExpert-IIT-B · Answer

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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a.(1/33)sigma(n=1 to n=16) {(4n^4)/((4n^2)-1)} = ? b.sigma(k=0 to k=n){(2^k)(2nCk)} = ?

a.(1/33)sigma(n=1 to n=16) {(4n^4)/((4n^2)-1)} = ?

b.sigma(k=0 to k=n){(2^k)(2nCk)} = ?

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