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 (n² + 5n + 4)

= n³ + 5n² + 4n

Therefore,

S_n = ∑ⁿ_{k = 1}(a)_k

= ∑ⁿ_{k = 1}(1/3k³ + 1/2k² + 1/6k)

= ∑ⁿ_{k = 1}(k)³ + 5∑ⁿ_{k = 1}(k)² + 4∑ⁿ_{k = 1}(k)

= n² (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 [(3n² + 3n + 20n + 10 + 24)/6]

= n (n + 1)/2 [(3n² + 23n + 34)/6]

= [n (n + 1)(3n² + 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)(3n² + 23n + 34)]/12