The sum of the following series 1 + 6 + 9 1 2 + 2 2 + 3 2 7 + 12 1 2 + 2 2 + 3 2 + 4 2 9 + 15 ( 1 2 + 2 2 +…

The sum of the following series 1+6+912+22+327+1212+22+32+429+15(12+22++52)11+.... up to 15 terms, is:
  1. 7520
  2. 7510
  3. 7830
  4. 7820

Solution

tn=3n12+22+32++n22n+1=3n×nn+12n+16 2n+1

=12 n3+n2

S15=n=115tn=n=11512n3+n2

=12n=115n3+n=115n2


=12×15×1622+15×16×316    n3=nn+122, n2=nn+1n+26

=7200+620

=7820

Asked in: JEE Main 2019 (09 Jan Shift 2)

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