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# Find the sum to n terms of the series whose n^{th} term is given by n (n + 1)(n + 4)

**Solution:**

The given n^{th} term is a_{n} = n (n + 1)(n + 4)

Hence,

a_{n} = n (n + 1)(n + 4)

= n (n^{2} + 5n + 4)

= n^{3} + 5n^{2} + 4n

Therefore,

S_{n} = ∑^{n}_{k = 1}(a)_{k}

= ∑^{n}_{k = 1}(1/3k^{3} + 1/2k^{2} + 1/6k)

= ∑^{n}_{k = 1}(k)^{3} + 5∑^{n}_{k = 1}(k)^{2} + 4∑^{n}_{k = 1}(k)

= n^{2} (n + 1)^{2}]/4 [5n (n + 1)(2n + 1)/6 + 4n (n + 1)/2]

= n (n + 1)/2 [n (n + 1)/2 + 5(2n + 1)/3 + 4]

= n (n + 1)/2 [(3n^{2} + 3n + 20n + 10 + 24)/6]

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

= [n (n + 1)(3n^{2} + 23n + 34)]/12

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 8

## Find the sum to n terms of the series whose n^{th} term is given by n (n + 1)(n + 4)

**Summary:**

It is known that a_{n} = n (n + 1)(n + 4) therefore S_{n} = ∑^{n}_{k = 1}(a)_{k}. So the sum to n terms of the series is [n (n + 1)(3n^{2} + 23n + 34)]/12

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