\( \dfrac{1^3}{1}+ \dfrac{1^3+2^3}{1+3}+ \dfrac{1^3+2^3+3^3}{1+3+5}+....\)

\( \dfrac{[\dfrac{n(n+1)}{2}]^2}{1+3+5+...+(2n-1)}\) is the nth term of the given series.

With first term a , last term (2n -1) and number of terms as n then 1,3,5...(2n -1) is an A.P.

Therefore,

= \( 1+3+5+...+(2n-1)\)

\( =\dfrac{n}{2}[2\times 1+(n-1)^2]=n^2\)

Also,

\( \dfrac{n^2(n+1)^2}{4n^2}=\dfrac{(n+1)^2}{4}\)

\( \dfrac{1}{4}n^2+\dfrac{1}{4}n+\dfrac{1}{4}\)

And,

\( \displaystyle\sum_{k=1}^{n}a_k=\displaystyle\sum_{k=1}^{n} (\dfrac{1}{4}k^2+\dfrac{1}{2}k+\dfrac{1}{4})\)

\( \dfrac{1}{4}\dfrac{n(n+1)(2n+1)}{6}+\dfrac{1}{2}\dfrac{n(n+1)}{2}+\dfrac{1}{4}n\)

= \( \dfrac{n[(n+1)(2n+1)+6(n+1)+6]}{24}\)

= \( \dfrac{n[2n^2+3n+1+6n+6+6]}{24}\)

=\( \dfrac{n[2n^2+9n+13]}{24}\)

Therefore, the sum of n terms of the given series

\( \dfrac{n[2n^2+9n+13]}{24}\)

Answered by Abhisek | 1 year 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 \( \dfrac{81}{2}\).