Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Asked by Abhisek | 1 year ago |  69

##### Solution :-

Given,

an = n (n + 1) (n + 4) = n(n+ 5n + 4) = n3 + 5n2 + 4n

Now, 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

$$s_n = 4\displaystyle\sum_{k=1}^{n}k^3+5\displaystyle\sum_{k=1}^{n}k^2 +4\displaystyle\sum_{k=1}^{n}k$$

$$\dfrac{n^2(n+1)^2}{4}+\dfrac{5n(n+1)(2n+1)}{6}$$

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

$$\dfrac{n(n+1)}{2}+[\dfrac{n(n+1)}{2} +$$

$$\dfrac{5n(n+1)}{3}+4]$$

$$\dfrac{n(n+1)}{2}$$

$$[\dfrac{3n^2+3n+20n+10+14}{6} ]$$

$$\dfrac{n(n+1)}{2}= [\dfrac{3n^2+23n+34}{6} ]$$

$$\dfrac{n(n+1)(3n^2+23n+34)}{12}$$

Therefore, the sum of nth terms of the series whose nth term is given by

$$\dfrac{n(n+1)(3n^2+23n+34)}{12}$$

Answered by Pragya Singh | 1 year ago

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