The value of the sum of 10 terms of the series \(S_{10}=\displaystyle\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+\ldots\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{10}{21}\)
Each term splits into two fractions whose middle parts cancel in pairs when the terms are added (a telescoping sum).
The general term (the denominators are \((2k)^2-1\) for k = 1, 2, 3, ...):
As a difference of squares, \((2k)^2-1=(2k-1)(2k+1)\)
Splitting into two fractions: \(\displaystyle\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\)
Adding the first 10 terms (k = 1 to 10):
Writing out the terms: \(\displaystyle S_{10}=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\ldots+\left(\frac{1}{19}-\frac{1}{21}\right)\right]\)
Everything cancels except the first and last parts: \(\displaystyle S_{10}=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}\times\frac{20}{21}=\frac{10}{21}\)