Given series is 3 ×12 + 5 × 22 + 7 × 32 + …
It’s seen that,
nth term, an = ( 2n + 1) n2 = 2n3 + n2
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 n terms of the given series is
\( S_n= \displaystyle\sum_{k=1}^{n}(2k^3+k^2)\)
= \( 2\displaystyle\sum_{k=1}^{n}k^3-\displaystyle\sum_{k=1}^{n}k^2\)
= \( 2[\dfrac{n(n+1)}{2}]^2+\dfrac{n(n+1)(2n+1)}{6}\)
= \( \dfrac{n^2(n+1)^2}{2}+\dfrac{n(n+1)(2n+1)}{6}\)
= \( \dfrac{n(n+1)}{2} [n(n+1)\dfrac{2n+1}{3}]\)
= \( \dfrac{n(n+1)}{2}[ \dfrac{3n^2+3n+2n+1}{3}]\)
= \( \dfrac{n(n+1)}{2} [ \dfrac{3n^2+5n+1}{3}]\)
= \( \dfrac{n(n+1)(3n^2+5n+1)}{6}\)
Therefore, the sum of n terms of the series is \( \dfrac{n(n+1)(3n^2+5n+1)}{6}\)
Answered by Pragya Singh | 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 \( \dfrac{81}{2}\).