Question

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

• A. 101
• B. 121
• C. 151
• D. does not exist

Answer 1

Answer: (C)

We have,
$\frac{2^{2}+4^{2}+6^{2}+\ldots+\left(2n\right)^{2}}{1^{2}+3^{2}+5^{2}+\ldots+\left(2n-1\right)^{2}}<\,1.01$
$=\frac{\Sigma4n^{2}}{\Sigma\left(4n^{2}-4n+1\right)}<\,1.01$
$=\frac{4\frac{n\left(n+1\right)\left(2n+1\right)}{6}}{4\frac{n\left(n+1\right)\left(2n+1\right)}{6}-4\frac{n\left(n\right)\left(n+1\right)}{2}+n}<\,1.01$
$=\frac{4\frac{n \left(n+1\right)\left(2n+1\right)}{6}}{\frac{n \left(2n+1\right)\left(2n-1\right)}{3}}<\, 101 = \frac{2\left(n+1\right)}{2n-1} <\, \frac{101}{100}$
$=200n+200<\,202n-101$
$\Rightarrow 2n>\,301$
$\Rightarrow n >\, \frac{301}{2}=150.5$
$\therefore n >\,151$