For frac{2^2+4^2+6^2+ldots+(2n)^2}{1^2+3^2+5^ | vedclass

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(D) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.We know that the numerator is $4(1^2+2^2+3^2+\ldots+n^2) = 4 	imes \frac{

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$150$

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(D) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.We know that the numerator is $4(1^2+2^2+3^2+\ldots+n^2) = 4 	imes \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.The denominator is the sum of squares of the first $n$ odd numbers,which is $\frac{n(2n-1)(2n+1)}{3}$.Thus,$S = \frac{2n(n+1)(2n+1)}{n(2n-1)(2n+1)} = \frac{2(n+1)}{2n-1}$.We are given $S > 1.01$,so $\frac{2n+2}{2n-1} > \frac{101}{100}$.Cross-multiplying,we get $200n + 200 > 202n - 101$.$301 > 2n$,which implies $n Since $n$ must be an integer,the maximum value of $n$ is $150$.

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

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