Question

Question 24 If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S2^2 = S3(1+8S1)

Class X1 - Maths -Sequences and Series Page 200

Answer 1

Here,
S₁ = sum of first n natural numbers
= n ( n + 1)/2 --------------------(1)

S₂ = sum of squares of first n natural numbers
= n( n + 1 )( 2n + 1 )/6 ---------(2)

S₃ = sum of cubes of first n natural numbers
= [ n( n + 1 )/2]²----------------(3)

now,
LHS = 9S₂²
= 9 [n( n + 1 )( 2n + 1 )/6 ]²
= 9[n²(n + 1)²(2n + 1)²/36 ]
= 9 × n²(n + 1)²/4 × (2n + 1)²/9
= [n(n + 1)/2]²{ 9 × (2n + 1)²/9}
= [n(n + 1)/2]²{ 4n² + 4n + 1}
= [n(n + 1)/2]²{1 + 4n(n + 1) }
= [n(n + 1)/2]²{ 1 + 8n(n+1)/2}
from equations (1) and (3)
= S₃ ( 1 + 8S₂ ) = RHS

Answer 2

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