Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Please,please help me - Permutations and combinations - JEE Main

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)

best_answer

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.

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question