The sum   \mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)} 

is equal to :

  • Option 1)

    620

  • Option 2)

    1240

  • Option 3)

    1860

  • Option 4)

    660

 

Answers (1)

 \mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)}

=\sum_{n=1}^{15}\frac{(\frac{n(n+1)}{2})^{2}}{\frac{n(n+1)}{2}}-\frac{1}{2}(\frac{15(1+15)}{2})

=\sum_{n=1}^{15}(\frac{n^{2}}{2}+\frac{n}{2})-\frac{1}{2}(\frac{15\times 16}{2})

=680-60

=620

So, option (1) is correct.

 


Option 1)

620

Option 2)

1240

Option 3)

1860

Option 4)

660

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