n^{th} term of the series as:

a_{n} = n^{2} + 2^{n}

Then, the sum of n terms of the series can be expressed as

\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)

The sum of first n terms of the given series is

\( S_n= \displaystyle\sum_{k=1}^{n}k^2+2^k\)

\( = \displaystyle\sum_{k=1}^{n}k^2+\displaystyle\sum_{k=1}^{n}2^k\)

Let \( \displaystyle\sum_{k=1}^{n}2^k=2^1+2^2+2^3+...\)

Both the first term and the common ratio of \( 2^1+2^2+2^3+...\) which forms a G.P. is 2 .

Therefore,

\( \displaystyle\sum_{k=1}^{n}2^k=\dfrac{(2)[(2)^n-1]}{2-1}\)

= \(2 (2^n-1)\)

Then,

\( S_n= \displaystyle\sum_{k=1}^{n}k^2+2(2^n+1)\)

\( \dfrac{n(n+1)(2n+1)}{6}+2n(n+1)n\)

Therefore, the sum of n terms of the series whose ntg term is given by \( n^2+2^n\)

is \( \dfrac{n(n+1)(2n+1)}{6}+2n(n+1)n\)

Answered by Pragya Singh | 2 years agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).