The sum of the following series 1+6+912+22+327+1212+22+32+429+1512+22+…+5211+… upto 15 terms, is

The sum of the following series 1+6+912+22+327+1212+22+32+429+1512+22++5211+ upto 15 terms, is

  1. A

    7510

  2. B

    7820

  3. C

    7830

  4. D

    7520

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    Solution:

    The nth  term of given expression is

    Tn=(3+(n1)3)12+22+32++n2(2n+1)

    =3nn(n+1)(2n+1)62n+1=n2(n+1)2

    Now, S15=12n=115n3+n2 =1215(15+1)22+15×16×316

    =12(120)2+1240=7820

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