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The minimum value of $\mathrm{n}$ for which $\frac{2^{2}+4^{2}+6^{2}+\ldots .+(2 \mathrm{n})^{2}}{1^{2}+3^{2}+5^{2}+\ldots .+(2 \mathrm{n}-1)^{2}} < 1.01$
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The correct answer is:
is 151
$\frac{x}{\frac{2 n(2 n+1)(4 n+1)}{6}-x} < 1.01$
$2.01 \mathrm{x} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$
$2.01 \cdot \frac{4 n(n+1)(2 n+1)}{6} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$
$\frac{2.01}{1.01} < \frac{4 n+1}{2 n+2} \Rightarrow \mathrm{n}>150.5$
$2.01 \mathrm{x} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$
$2.01 \cdot \frac{4 n(n+1)(2 n+1)}{6} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$
$\frac{2.01}{1.01} < \frac{4 n+1}{2 n+2} \Rightarrow \mathrm{n}>150.5$
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