8. sequence and series — MATHS STD 11 Science — Question

We have,

$\frac{2^2+4^2+6^2+\ldots+(2 n)^2}{1^2+3^2+5^2+\ldots+(2 n-1)^2} < 1.01$

$=\frac{\Sigma 4 n^2}{\sum\left(4 n^2-4 n+1\right)} < 1.01$

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

$= \frac{4 \frac{n(n+1)(2 n+1)}{6}}{\frac{n(2 n+1)(2 n-1)}{3}}<101=\frac{2(n+1)}{2 n-1} < \frac{101}{100}$

$= 200 n+200 < 202 n-101$

$\Rightarrow 2 n > 301 \Rightarrow n > \frac{301}{2}=150.5$

$\therefore n > 151$