Given series, (43)3+(46)3+(49)3+(412)3+(415)3+…15 terms
=(43)3(13+23+33+…+153)
=6427(215×16)2[∵n=1∑n=nn3=[2n(n+1)]2]
=6427×1225×64=27×225
Since, given 225K=27×225
⇒K=27
JEE Main 2019 — Mathematics Algebra
If the sum of the first 15 terms of the series (43)3+(121)3+(241)3+33+(343)3+… is equal to 225K, then K is equal to :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
9
27
54
108
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