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The sum of the squares of the first n odd natural number is given by;

$S_n=\frac{n(2n+1)(2n-1)}{3}$

It is given that $S_n=533n$ so;

$\implies533n*3=n((2n)^2-1)$

$\implies 1599*n=4n^3-n$

$\implies 1599*n=n(4n^2-1)$

$\implies 1599+1=4n^2$

$\implies 1600=4n^2$

$\implies n^2=\frac{1600}{4}$

$\implies n^2=400$

$\implies n=20$

Option (B) is correct.

$S_n=\frac{n(2n+1)(2n-1)}{3}$

It is given that $S_n=533n$ so;

$\implies533n*3=n((2n)^2-1)$

$\implies 1599*n=4n^3-n$

$\implies 1599*n=n(4n^2-1)$

$\implies 1599+1=4n^2$

$\implies 1600=4n^2$

$\implies n^2=\frac{1600}{4}$

$\implies n^2=400$

$\implies n=20$

Option (B) is correct.