For $\frac{2^{2} + 4^{2} + 6^{2} + . . . . (2 n)^{2}}{1^{2} + 3^{2} + 5^{2} + . . . . + (2 n - 1)^{2}}$ to exceed 1.01,the maximum value of n is

99

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100

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101

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150

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Solution

The correct option is D $150$
$\frac{2^{2} [1^{2} + 2^{2} + 3^{2} + . . . . + n^{2}]}{[1^{2} + 2^{2} + 3^{2} + . . . . + {(2 n)}^{2}] - 2^{2} [1^{2} + 2^{2} + 3^{2} + . . . . + n^{2}]}$
$= \frac{2^{2} \frac{n (n + 1) (2 n + 1)}{6}}{\frac{2 n (2 n + 1) (4 n + 1)}{6} - \frac{2^{2} n (n + 1) (2 n + 1)}{6}} > 1.0$
$\Rightarrow \frac{2 n + 2}{2 n - 1} > 1.01$
$\Rightarrow 1 + \frac{3}{2 n - 1} > 1.01$
$\Rightarrow 2 n - 1 > 300$
$\Rightarrow n < 150.5$

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Similar questions

\frac{2^{2} + 4^{2} + 6^{2} + . . . . + (2 n)^{2}}{1^{2} + 3^{2} + 5^{2} + . . . . . + (2 n - 1)^{2}}

1.01

?

\frac{2^{2} + 4^{2} + 6^{2} + . . . . . + (2 n)^{2}}{1^{2} + 3^{2} + 5^{2} + . . . . + (2 n - 1)^{2}} < 1.01

Write the sum of the series : $1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . + (2 n - 1)^{2} - (2 n)^{2}$

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . . . . . .

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