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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $\frac{\left[ \frac{n(n+1)}{2} ight]^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2}{4} 	imes \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2(2n+1)}$.Substituting $n = 12$:Ratio $= \frac{3 	imes 12 	imes (12+1)}{2 	imes (2 	imes 12 + 1)} = \frac{3 	imes 12 	imes 13}{2 	imes 25} = \frac{3 	imes 6 	imes 13}{25} = \frac{234}{25}$.

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

Similar Questions

Maximum value of the sum of the Arithmetic Progression $50, 48, 46, 44, \dots$ is

The $5^{th}$ term of a $G.P.$ is $2,$ find the product of first $9$ terms.

If the ${p^{th}}$ term of an $A.P.$ is $q$ and the ${q^{th}}$ term is $p$,then its ${r^{th}}$ term will be

If the numbers $a, b, c, d, e$ form an $A.P.$,then the value of $a - 4b + 6c - 4d + e$ is

Find the sum of infinite terms of the series $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module