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If $S_1$ and $S_2$ are the variances of the first $2 k$ and $k(k>1)$ natural numbers respectively, then $\left(S_1 / S_2\right)$ lies in the interval
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$(1,4]$
The variance of first $2 k$ natural numbers
$S_1=\frac{2 k(2 k+1)(4 k+1)}{6 \times 2 k}-\left(\frac{2 k(2 k+1)}{2 \times 2 k}\right)^2$
$=(2 k+1)\left[\frac{4 k+1}{6}-\frac{2 k+1}{4}\right]$
$=\frac{2 k+1}{12}[8 k+2-6 k-3]=\frac{4 k^2-1}{12}$
and variance of first $k$ natural numbers
$S_2=\frac{k(k+1)(2 k+1)}{6 \times k}-\left(\frac{k(k+1)}{2 \times k}\right)^2$
$=(k+1)\left[\frac{2 k+1}{6}-\frac{k+1}{4}\right]$
$=\frac{k+1}{12}[4 k+2-3 k-3]=\frac{k^2-1}{12}$
$\therefore \quad \frac{S_1}{S_2}=\frac{4 k^2-1}{k^2-1}=4+\frac{3}{k^2-1},(k>1) \quad \therefore \frac{S_1}{S_2} \in(1,4]$
$S_1=\frac{2 k(2 k+1)(4 k+1)}{6 \times 2 k}-\left(\frac{2 k(2 k+1)}{2 \times 2 k}\right)^2$
$=(2 k+1)\left[\frac{4 k+1}{6}-\frac{2 k+1}{4}\right]$
$=\frac{2 k+1}{12}[8 k+2-6 k-3]=\frac{4 k^2-1}{12}$
and variance of first $k$ natural numbers
$S_2=\frac{k(k+1)(2 k+1)}{6 \times k}-\left(\frac{k(k+1)}{2 \times k}\right)^2$
$=(k+1)\left[\frac{2 k+1}{6}-\frac{k+1}{4}\right]$
$=\frac{k+1}{12}[4 k+2-3 k-3]=\frac{k^2-1}{12}$
$\therefore \quad \frac{S_1}{S_2}=\frac{4 k^2-1}{k^2-1}=4+\frac{3}{k^2-1},(k>1) \quad \therefore \frac{S_1}{S_2} \in(1,4]$
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