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The value of the following sum of terms $$\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\frac{9}{4^2 \cdot 5^2}+\frac{11}{5^2 \cdot 6^2}+\frac{13}{6^2 \cdot 7^2}+\frac{15}{7^2 \cdot 8^2}$$
The nth term of the series can be written as:
$$\frac{2n+1}{n^2\left(n+1\right)^2}$$.
Now, 2n+1 can be written as: $$\left(n+1\right)^2-n^2$$.
This gives nth term to be:
$$\frac{\left(\left(n+1\right)^2-n^2\right)}{n^2\times\ \left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}$$.
Now, we put values of n from 2 to 7 and add them to get:
$$\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...\left(\frac{1}{7^2}-\frac{1}{8^2}\right)=\frac{1}{2^2}-\frac{1}{8^2}$$.
==> $$\frac{1}{4}-\frac{1}{64}=\frac{15}{64}$$.
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