tn=2n+13n(12+22+32+…+n2)=6(2n+1)3n×n(n+1)(2n+1)
=21(n3+n2)
∴S15=n=1∑15tn=n=1∑1521(n3+n2)
=21[n=1∑15n3+n=1∑15n2]
=21×[(215×16)2+615×16×31] ∑n3=(2n(n+1))2,∑n2=(6n(n+1)(n+2))
=7200+620
=7820
JEE Main 2019 — Mathematics Algebra
The sum of the following series 1+6+79(12+22+32)+912(12+22+32+42)+1115(12+22+…+52)+.... up to 15 terms, is:
Held on 9 Jan 2019 · Verified 6 Jul 2026.
7520
7510
7830
7820
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