Consider 12+32+52+……+252 nth term Tn=(2n−1)2,n=1,……13 Now, Sn=n=1∑13 Tn=n=1∑13(2n−1)2 =n=1∑134n2+n=1∑131−n=1∑134n=4∑n2+13−4∑n =4[6n(n+1)(2n+1)]+13−42n(n+1) Put n=13, we get Sn=26×14×9+13−26×14=3276+13−364=2925.
JEE Main 2013 — Mathematics Algebra

Held on 25 Apr 2013 · Verified 6 Jul 2026.
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