frac{1^3 + 2^3 + 3^3 + 4^3 + dots + 12^3}{1^2 | vedclass

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Question

(A) The sum of the first $n$ cubes is given by $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2$. The sum of the first $n$ squares is given by

Vedclass · Accepted Answer

$\frac{234}{25}$

Vedclass · Answer

(A) The sum of the first $n$ cubes is given by $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} ight]^2$. The sum of the first $n$ squares is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. Therefore,the ratio is: $\frac{\sum_{k=1}^{12} k^3}{\sum_{k=1}^{12} k^2} = \frac{\left[ \frac{n(n+1)}{2} ight]^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2}{4} 	imes \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2(2n+1)}$. Substituting $n = 12$: $\frac{3 	imes 12 	imes (12+1)}{2 	imes (2 	imes 12 + 1)} = \frac{3 	imes 12 	imes 13}{2 	imes 25} = \frac{3 	imes 6 	imes 13}{25} = \frac{234}{25}$.

$\frac{1^3 + 2^3 + 3^3 + 4^3 + \dots + 12^3}{1^2 + 2^2 + 3^2 + 4^2 + \dots + 12^2} = $

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