The minimum value of n for which frac{2^2+4^2 | vedclass

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(C) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.The numerator is $\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = 4 \frac{n

Vedclass · Accepted Answer

$151$

Vedclass · Answer

(C) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.The numerator is $\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = 4 \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.The denominator is $\sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n = \frac{n(2n+1)(2n-1)}{3}$.Thus,$S = \frac{2n(n+1)(2n+1)/3}{n(2n+1)(2n-1)/3} = \frac{2(n+1)}{2n-1}$.We are given $S $100(2n+2) $301  150.5$.Since $n$ must be an integer,the minimum value is $n = 151$.

The minimum value of $n$ for which $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2} < 1.01$ is

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