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 square plus 2 raised to n

9.   Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n^2 + 2 ^ n

Answers (1)

best_answer

nth terms are given by n^2 + 2 ^ n

a_n=n^2 + 2 ^ n

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

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

This term is a GP with first term =a =2 and common ratio =r =2.

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

 

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

     S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)

Thus, the sum is 

                        S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)

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

SC seeks CBSE reply on Saudi-Based student's plea over withheld class 12 improvement result
anam-khan

CBSE explains 'roll number not found' error on portal amid student concerns
anam-khan

Congress attacks Dharmendra Pradhan over CBSE row, alleges failures in exam and revaluation process

Latest Question