Given series is 3 × 8 + 6 × 11 + 9 × 14 + …
It’s found out that,
an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)
= (3n) (3n + 5)
= 9n2 + 15n
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
\( S_n= \displaystyle\sum_{k=1}^{n}(9k^2+15k)\)
\( = 9\displaystyle\sum_{k=1}^{n}k^2+15\displaystyle\sum_{k=1}^{n}k\)
\(9\times \dfrac{n(n+1)(2n+1)}{6}+15\times \dfrac{n(n+1)}{2}\)
\( \dfrac{3n(n+1)(2n+1)}{2}\dfrac{15n(n+1)}{2}\)
= \( \dfrac{3n(n+1)}{2}(2n+1+5)\)
= \( \dfrac{3n(n+1)}{2}(2n+6)\)
= \( 3n(n+1)(n+3)\)
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 \( \dfrac{81}{2}\).