If S₁, S₂, S₃ are the sums of first n natural numbers, their squares and their cubes respectively, show that 9S₂² = S₃ (1 + 8S₁)

Solution:

Using the summation formulas,

S₁ = n(n + 1)/2

S₂ = n (n + 1)(2n + 1)/6

S₃ = [n(n + 1)/2]²

Hence,

9S₂² = 9 [n (n + 1)(2n + 1)/6]²

= 9/36 [n (n + 1)(2n + 1)]²

= 1/4 [n (n +1 )(2n + 1)]²

= [n (n + 1)(2n + 1)/2]² ....(1)

Also,

S₃ (1 + 8S₁) = [n(n + 1)/2]² x [1 + 8n(n + 1)/2]

= [n(n + 1)/2]² x [1+ 4n² + 4n]

= [n (n + 1)(2n + 1)/2]² ....(2)

Thus, from (1) and (2), we obtain

9S₂² = S₃ (1 + 8S₁)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 24

If S₁, S₂, S₃ are the sums of first n natural numbers, their squares and their cubes respectively, show that 9S₂² = S₃ (1 + 8S₁).

Summary:

If S₁, S₂, S₃ are the sums of first n natural numbers, their squares and their cubes respectively, we have shown that 9S₂² = S₃ (1 + 8S₁)