The number of positive integers n in the set | vedclass

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Question

(B) We have the expression $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$.Using the standard summation formulas,the numerator is $\frac{n(n+1)(2n+1)}

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(B) We have the expression $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$.Using the standard summation formulas,the numerator is $\frac{n(n+1)(2n+1)}{6}$ and the denominator is $\frac{n(n+1)}{2}$.Dividing these,we get $\frac{n(n+1)(2n+1)}{6} 	imes \frac{2}{n(n+1)} = \frac{2n+1}{3}$.For this expression to be an integer,let $\frac{2n+1}{3} = k$,where $k$ is an integer.This implies $2n+1 = 3k$,or $2n = 3k-1$.Since $n$ must be an integer,$3k-1$ must be even,which means $k$ must be odd.Given $1 \leq n \leq 100$,we have $1 \leq \frac{3k-1}{2} \leq 100$.$2 \leq 3k-1 \leq 200$ $\Rightarrow 3 \leq 3k \leq 201$ $\Rightarrow 1 \leq k \leq 67$.We need to count the number of odd integers $k$ in the range $[1, 67]$.The odd integers are $1, 3, 5, \ldots, 67$.This is an arithmetic progression with $a=1$,$d=2$,and $l=67$.Using $l = a + (m-1)d$,we get $67 = 1 + (m-1)2$ $\Rightarrow 66 = 2(m-1)$ $\Rightarrow 33 = m-1$ $\Rightarrow m = 34$.Thus,there are $34$ such integers $n$.

The number of positive integers $n$ in the set $\{1, 2, 3, \ldots, 100\}$ for which the number $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$ is an integer is

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