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  • 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

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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

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