Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n (n+1) (n+4).

8.   Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n (n+1) ( n + 4 )

Answers (1)

best_answer

nth terms is given by n (n+1) ( n + 4 )

a_n=n (n+1) ( n + 4 )=n(n^2+5n+4)=n^3+5n^2+4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)(k+4)

                        =\sum _{k=1}^{n} k^3+5\sum _{k=1}^{n} k^2+4\sum _{k=1}^{n} k

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+\frac{4.n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+2.n(n+1)

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+20n+10+24}{6} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+23n+34}{6} \right )

                =\left [ \frac{n(n+1)}{24} \right ] \left ( 3n^2+23n+34 \right )

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 Board Exam 2026 LIVE: 12th Accountancy set 1, 2 question papers with solutions; answer key
anam-khan

CBSE Class 12 Physics paper well-balanced, moderate in difficulty; MCQ lengthy, calculation-based
anam-khan

CBSE Board 2026 Exam LIVE: Class 12 physics 'moderate' in difficulty; set-wise answer key updates

Latest Question