Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + ....

Solution:

The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + .... n terms.

a_n = (1² + 2² + 3² + .... + n²)

= n/6 (n + 1)(2n + 1)

= n/6 (2n² + 3n + 1)

= 1/3 n³ + 1/2 n² + 1/6 n

Therefore,

S_n = ∑ⁿ_{k = 1}(a)_k

= ∑ⁿ_{k = 1}(1/3k³ + 1/2k² + 1/6k)

= 2∑ⁿ_{k = 1}(k)³ + ∑ⁿ_{k = 1}(k)²

= 1/3[n² (n + 1)²]/2 + 1/2[n (n + 1)(2n + 1)]/6 + 1/6 [n (n + 1)]/2

= n (n + 1)/6 [n (n + 1)/2 + (2n + 1)/2 + 1/2]

= n (n + 1)/6 [n² + n + 2n + 1 + 1]/2

= n (n + 1)/6 [n² + n + 2n + 2]/2

= n (n + 1)/6 [n (n + 1) + 2 (n + 1)]/2

= n (n + 1)/6 [(n + 1)(n + 2)]/2

= [n (n + 1)² (n + 2)]/12

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 7

Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + ....

Summary:

As we can see that this was an increasing series and therefore the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + .... is [n (n + 1)² (n + 2)]/12