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: $\frac{\sum_{k=1}^{12} k^3}{\sum_{k=1}^{12} k^2} = \frac{\left[ \frac{n(n+1)}{2} ight]^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2}{4} 	imes \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2(2n+1)}$. 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} = $

Similar Questions

If the set of natural numbers is partitioned into subsets $S_1 = \{1\}, S_2 = \{2, 3\}, S_3 = \{4, 5, 6\}$ and so on,then the sum of the terms in $S_{50}$ is

The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

$2.\overline{357} = $

For all $n \in N$,if $1^3+2^3+3^3+\ldots+n^3 > x$,then a value of $x$ among the following is

Vedclass Test Series

Exam Paper Generator

Online Exam Module