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 2n minus 1 square

10.   Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by  ( 2n-1) ^2

Answers (1)

best_answer

nth terms is given by  ( 2n-1) ^2.

a_n=( 2n-1) ^2=4n^2+1-4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (2k-1)^2

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

                     =\frac{4.n(n+1)(2n+1)}{6}-\frac{4.n(n+1)}{2}+n

                   =\frac{2.n(n+1)(2n+1)}{3}-2.n(n+1)+n

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

                   =n(\frac{4n^2+6n+2-6n-6+3}{3})

                     =n(\frac{4n^2-1}{3})

                     =n(\frac{(2n+1)(2n-1)}{3}) 

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 exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question