The value of \sum_r=1^15r^2\, \left ( \frac^15\textrm \right ) is equal to:

  • Option 1)

     560

  • Option 2)

    680

  • Option 3)

    1240

  • Option 4)

     1085

     

 

Answers (1)

As we learnt in

Theorem of Combination -

Each of the different groups or selection which can be made by taking r things from n things is called a combination.

^{n}c_{r}=\frac{(n)!}{r!(n-r)!}

- wherein

Where 1\leq r\leq n

 

 Special series of sequences and series. 

\Rightarrow\ \;\sum_{r=1}^{15}r^{2}\times \frac{\frac{15!}{r!15-r!}}{\frac{15!}{(r-1)!(16-r)!}}

\Rightarrow\ \;\sum_{r=1}^{15}r^{2}\times \frac{(16-r)}{r}

\Rightarrow\ \;\sum_{r=1}^{15}r(16-r)

\Rightarrow\ \;\sum_{r=1}^{15}16r-\sum_{r=1}^{15}16r^{2}

\Rightarrow\ \;\frac{16\times15\times16}{2}-\frac{15\times16\times31}{6}=1920-1240=680

Correct option is 2.

 

 


Option 1)

 560

This is an incorrect option.

Option 2)

680

This is the correct option.

Option 3)

1240

This is an incorrect option.

Option 4)

 1085

 

This is an incorrect option.

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