The sum of all positive integers n for which | vedclass

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

Question

(A) The given expression is $S = \frac{\sum_{k=1}^{2n} k^3}{\sum_{k=1}^{n} k^2}$.Using the formulas $\sum_{k=1}^{m} k^3 = \left(\frac{m(m+1)}{2}\right

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The given expression is $S = \frac{\sum_{k=1}^{2n} k^3}{\sum_{k=1}^{n} k^2}$.Using the formulas $\sum_{k=1}^{m} k^3 = \left(\frac{m(m+1)}{2}\right)^2$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we have:$S = \frac{\left(\frac{2n(2n+1)}{2}\right)^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(2n+1)^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{6n^2(2n+1)^2}{n(n+1)(2n+1)} = \frac{6n(2n+1)}{n+1}$.Now,simplify the expression:$S = \frac{12n^2+6n}{n+1} = \frac{12n(n+1) - 6n}{n+1} = 12n - \frac{6n}{n+1} = 12n - \frac{6(n+1)-6}{n+1} = 12n - 6 + \frac{6}{n+1}$.For $S$ to be an integer,$n+1$ must be a divisor of $6$.The divisors of $6$ are $1, 2, 3, 6$.Thus,$n+1 \in \{1, 2, 3, 6\}$,which gives $n \in \{0, 1, 2, 5\}$.Since $n$ must be a positive integer,$n \in \{1, 2, 5\}$.The sum of these values is $1+2+5 = 8$.

The sum of all positive integers $n$ for which $\frac{1^3+2^3+\ldots+(2n)^3}{1^2+2^2+\ldots+n^2}$ is also an integer is

Similar Questions

Let $a_n = (1^2 + 2^2 + \ldots + n^2)^n$ and $b_n = n^n(n!)$. Then

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

Let $\{a_{n}\}_{n=0}^{\infty}$ be a sequence such that $a_{0}=a_{1}=0$ and $a_{n+2}=2a_{n+1}-a_{n}+1$ for all $n \geq 0$. Then,$\sum\limits_{n=2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to

If the sum of the squares of the first $n$ natural numbers exceeds their sum by $330$,then $n = $

Find the sum of the sequence $7, 77, 777, 7777, \ldots$ to $n$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module