# If S₁, S₂, S₃ are the sums of first n natural numbers, their squares and their cubes respectively, show that 9S₂^{2} = 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,

9S₂^{2} = 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+ 4n^{2} + 4n]

= [n (n + 1)(2n + 1)/2]^{2} ....(2)

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

9S₂^{2} = S₃ (1 + 8S₁)

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 9S₂^{2} = 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 9S₂^{2} = S₃ (1 + 8S₁)