Question

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

• A. $99$
• B. $100$
• C. $101$
• D. $150$

Answer 1

The correct option is D $150$

$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Also, $1^2+2^2+...+(2n{)}^2=\frac{2n(2n+1)(4n+1)}{6}$

Let the numerator be $X$ and the denominator be denoted by $Y$

$X+Y=\frac{2n(2n+1)(4n+1)}{6}$

Also, $X=4(1^2+2^2+...+n^2)=\frac{4n(n+1)(2n+1)}{6}$

$\Rightarrow Y=\frac{2n(2n+1)(4n+1)}{6}-\frac{4n(n+1)(2n+1)}{6}$

$=\frac{2n(2n+1)(2n-1)}{6}$

$\Rightarrow \frac{X}{Y}=\frac{4n(n+1)(2n+1)}{2n(2n+1)(2n-1)}$

$=\frac{2n+2}{2n-1}$

According to the given information, $\frac{2n+2}{2n-1}>1.01$

$\Rightarrow 100(2n+2)>101(2n-1)$

$\Rightarrow 200n+200>202n-101$

$\therefore 2n<301$

or $n<150.5$

Maximum value of $n$ is thus $150$