Question

In the given expression $\frac{3}{1^2{.2}^2}+\frac{5}{2^2{.3}^2}+\frac{7}{3^2{.4}^2}+\frac{9}{4^2{.5}^2}+\frac{11}{5^2{.6}^2}+\frac{13}{6^2{.7}^2}+\frac{15}{7^2{.8}^2}+\frac{17}{8^2{.9}^2}+\frac{19}{9^2{.10}^2}=\frac{99}{x}$. Find the value of $x$

Answer 1

Given exp. = $(1-\frac{1}{2^2})+(\frac{1}{2}-\frac{1}{3^2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5^2}-\frac{1}{6^2})+(\frac{1}{6^2}-\frac{1}{7^2})+$
$(\frac{1}{7^2}-\frac{1}{8^2})+(\frac{1}{8^2}-\frac{1}{9^2})+(\frac{1}{9^2}-\frac{1}{{10}^2})$
= 1 - $\frac{1}{{10}^2}=1-\frac{1}{100}=\frac{99}{100}$

The value is $100$