Find the sum of 1 3 /1+ (1 3 +2 3 /1+3) +(1 3 +2 3 +3 3 /1+3+5) + ….......16 terms. Options: 442,446,448,447

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Arun · Answer

Dear Neetu For nth term Tn = (1&sup3; + 2&sup3; +3&sup3;+..,..+n&sup3;)/(1 +3+5+....(2n-1)) Tn = n&sup2;(n+1)&sup2;/[4 *(n/2)(2n)] Tn = (n+1)&sup2;/4 Tn = (n&sup2; + 1 + 2n) /4 Now Sn = sigma Tn Then put n = 16 in Sn Now you can solve Regards Arun (askIITians forum expert)

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Find the sum of 1 3 /1+ (1 3 +2 3 /1+3) +(1 3 +2 3 +3 3 /1+3+5) + ….......16 terms. Options: 442,446,448,447

Find the sum of 1³/1+ (1³+2³/1+3) +(1³+2³+3³/1+3+5) + ….......16 terms.

Options: 442,446,448,447

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Find the sum of 1 3 /1+ (1 3 +2 3 /1+3) +(1 3 +2 3 +3 3 /1+3+5) + ….......16 terms. Options: 442,446,448,447

Find the sum of 13/1+ (13+23/1+3) +(13+23+33/1+3+5) + ….......16 terms. Options: 442,446,448,447

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Find the sum of 1³/1+ (1³+2³/1+3) +(1³+2³+3³/1+3+5) + ….......16 terms.

Options: 442,446,448,447