Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Asked by Abhisek | 1 year ago |  92

##### Solution :-

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

a= (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 ago

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