Find the sum to n terms of the series 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + 4 x 5 + ....

Solution:

The given series is 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + 4 x 5 + .... n terms.

Hence,

a_n = n (n + 1)(n + 2)

= (n² + n)(n + 2)

= n³ + 3n² + 2n

Therefore,

S_n = ∑ⁿ_{k = i}(a)_k

= ∑ⁿ_{k = 1}(k³ + 3k² + 2k)

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

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

= [n (n + 1)/2]² + [n (n + 1)(2n + 1)/2] + (n + 1)/2

S_n = n (n + 1)/2 [n (n + 1)/2 + 2n + 1 + 2]

= n (n + 1)/2 [(n² + n + 4n + 6)/2]

= n (n + 1)/4 [n² + 5n + 6]

= n (n + 1)/4 [n² + 2n + 3n + 6]

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

= n/4 (n + 1)(n + 2)(n + 3)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 2

Find the sum to n terms of the series 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + 4 x 5 + ....

Summary:

The given series is 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + 4 x 5 + .... up to n terms was found out to be n/4 (n + 1)(n + 2)(n + 3) using a_n = n (n + 1)(n + 2)