Find the sum to n terms of the series 5² + 6² + 7² + .... + 20²

Solution:

The given series is 5² + 6² + 7² + .... + 20²

Hence,

a_n = (n + 4)²

= n² + 8n + 16

Therefore,

S_n = ∑ⁿ_{k = 1}(a)_k

= ∑ⁿ_{k = 1}(k² + 8k + 16)

= 2∑ⁿ_{k = 1}(k)² + 8∑ⁿ_{k = 1}k + ∑ⁿ_{k = 1}16

= [n (n + 1)(2n + 1)]/6 + [8n (n + 1)]/2 + 16n

Since, 16^th term is (16 + 4)² = (20)²

Then,

S₁₆ = [16(16 + 1)(2(16) + 1)]/6 + [8(16)(16 + 1)]/2 + 16 (16)

= [(16)(17)(33)]/6 + [(128)(17)]/2 + 256

= 1496 + 1088 + 256

= 2840

Thus, 5² + 6² + 7² + .... + 20² is 2840

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 5

Find the sum to n terms of the series 5² + 6² + 7² + .... + 20²

Summary:

We have the series 5² + 6² + 7² + .... + 20². And we know that S_n = ∑ⁿ_{k = 1}(a)_k. Therefore the sum of these terms is 5² + 6² + 7² + .... + 20² is 2840