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

What term of the progression $18, -12, 8, \ldots$ is $\frac{512}{729}$?

The $4^{th}$ term of a $H.P.$ is $\frac{3}{5}$ and $8^{th}$ term is $\frac{1}{3},$ then its $6^{th}$ term is

Three numbers are selected from the set ${3^1, 3^2, 3^3, \dots, 3^{20}}$. Find the number of ways these selected numbers can form an increasing Geometric Progression $(G.P.)$.

The $n^{th}$ term of the series $3.8 + 6.11 + 9.14 + 12.17 + .....$ will be

If $n$ is odd or even,then the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module