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

$\frac{2^2[1^2+2^2+...+n^2]}{[1^2+3^2+5^2+...+(2n-1{)}^2]}$
$1^2+2^2+3^2+....+(2n{)}^2=\frac{2n(2n+1)(4n+1)}{6}$
$[1^2+3^2....+(2n-1{)}^2+2^2[1^2+2^n.....+n^2]$
$=\frac{2n(2n+1)(4n+1)}{6}$
$S+4\frac{n(n+1)(2nn+1)}{6}=\frac{2n(2n+1)(4n+1)}{6}$
$S+\frac{2n(2n+1)(4n+1)}{6}-\frac{4n(n+1)(2n+1)}{6}$
$=2n\left(\frac{2n+1}{6}\right)[4n+1-2n-2]$
$=\frac{2n(2n+1)(2n-1)}{6}$
$Ratio=\frac{4n(n+1)(2n+1)}{6}\times \frac{6}{2n(2n+1)(2n-1)}=\frac{2n+2}{2n-1}$
$\frac{2n+2}{2n-1}>\frac{101}{100}$
$200n+200>202n-101$
$2n<301$
$n<\frac{301}{2}\Rightarrow$ maximum value $=150$