Given sum is 123+12+225+12+22+327+…..nth term =Tn=6n(n+1)(2n+1)2n+1=n(n+1)6 or Tn=6[n1−n+11]∴Sn=∑Tn=6∑n1−6∑n+11=n6n−n+16=6−n+16=n+16nS11=11+16×11=1266=633=211
The sum 3 1^2+ 5 1^2+2^2+ 7 1^2+2^2+3^2+…. upto 11-terms is:
JEE Main 2013 — Mathematics Algebra
The sum 123+12+225+12+22+327+…. upto 11-terms is:
Held on 22 Apr 2013 · Verified 6 Jul 2026.
Options
- A
27
- B
411
- C
211
- D
1160
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