Question

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

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

Answer 1

Answer: (C)

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