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 each of the series 3 × 1^2 + 5 × 2^2 + 7 × 3^2 + ...

3.   Find the sum to n terms of each of the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

Answers (1)

best_answer

the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

nth term  = (2n+1)(n^2)=2n^3+n^2=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} 2k^3+k^2

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

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

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

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

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

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

                  = \frac{n(n+1)(3n^2+5n+1)}{6}   

Thus, the sum is 

                       = \frac{n(n+1)(3n^2+5n+1)}{6}

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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question