Question
$\frac{1^2+2^2+\hat{\jmath}^2+\ldots \ldots . .+n^2}{1+2+\hat{\jmath}+\ldots \ldots \ldots+n}$ के पूर्णांक $(integer)$ होने के लिए $\{1,2,3, \ldots, 100)$ समुच्चय (set) में कितने धनात्मक पूर्णाक $n$ होंगे ?
We have,
$\frac{1^2+2^2+3^2+4^2+\ldots+n^2}{1+2+3+4+\ldots+n}$
$=\frac{\frac{n(n+1)(2 n+1)}{6}}{n(n+1)}$
$=\frac{2 n+1}{2}=k(l \text { let })$
$\Rightarrow \quad n=\frac{3 k-1}{2}$
$\text { Now, } 1 \leq \frac{3 k-1}{2} \leq 100$
$\Rightarrow \quad 2 \leq 3 k-1 \leq 200$
$\Rightarrow \quad 2+1 \leq 3 k \leq 200+1$
$\Rightarrow \quad 3 \leq 3 k \leq 201$
$\Rightarrow \quad 1 \leq k \leq 201$
$\Rightarrow \quad 1 \leq k \leq 67$
Number of odd integer $=34$
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