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.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions