1³/1 +1³+2³/1+2 + 1³+2³+3³/1+2+3........... Sum of n terms
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Tn = (13 + 23 + 33 + ………)/(1 + 3 + 5 ……..)n times = {n2(n + 1)2/4}/{(n/2)(2 + 2n – 2)} So, Tn = {n2(n + 1)2/4}/(n2) Tn = (n + 1)2/4 So, Sn = (1/4) [ n(n + 1)(2n + 1)/6 + n + 2n(n + 1)/2 ] Sn = (n/4) [ (n + 1)(2n + 1)/6 + 1 + (n + 1) ] Sn = (n/24) [ (2n2 + 3n + 1) + 6n + 12 ] Sn = (n/24) [ 2n2 + 9n + 13 ]
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