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If $S_1, S_2, S_3$ denote the sums of the first $n$ natural numbers, their squares, and their cubes respectively, which identity is correct?
A$9S_2^2=S_3(1+8S_1)$
B$S_2^2=9S_3(1+8S_1)$
C$9S_3^2=S_2(1+8S_1)$
D$S_3^2=9S_2(1+8S_1)$
Answer & Solution
Correct answer: A. $9S_2^2=S_3(1+8S_1)$
Use
$$S_1=\frac{n(n+1)}{2},\quad S_2=\frac{n(n+1)(2n+1)}{6},\quad S_3=\frac{n^2(n+1)^2}{4}.$$Then
$$9S_2^2=9\left(\frac{n(n+1)(2n+1)}{6}\right)^2=\frac{1}{4}n^2(n+1)^2(2n+1)^2.$$Also,
$$S_3(1+8S_1)=\frac{1}{4}n^2(n+1)^2\left(1+8\cdot \frac{n(n+1)}{2}\right)=\frac{1}{4}n^2(n+1)^2(2n+1)^2.$$So both sides are equal, proving option A.