\( \dfrac{1^3}{1}+ \dfrac{1^3+2^3}{1+3}+ \dfrac{1^3+2^3+3^3}{1+3+5}+....\)
\( \dfrac{[\dfrac{n(n+1)}{2}]^2}{1+3+5+...+(2n-1)}\) 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).