Question Bank › General Aptitude › Simplification › ABC26GN1073ABC26GN1073 · Simplification Subject: General Aptitude · Chapter: Simplification · Exam: · Marks: · Difficulty:
What is the value of the following expression? $\frac{1}{\left(2^{2}-1\right)}+\frac{1}{\left(4^{2}-1\right)}+\frac{1}{\left(6^{2}-1\right)}+\cdots+\frac{1}{\left(20^{2}-1\right)}$
(a) $\frac{9}{19}$
(b) $\frac{10}{19}$
(c) $\frac{10}{21}$
(d) $\frac{11}{21}$
Answer Explanation Given $\exp .=\frac{1}{(2-1)(2+1)}+\frac{1}{(4-1)(4+1)}$ $$+\frac{1}{(6-1)(6+1)}+\ldots . .+\frac{1}{(20-1)(20+1)} \begin{aligned} & =\frac{1}{1 \times 3}+\frac{1}{3 \times 5}+\frac{1}{5 \times 7}+\ldots . .+\frac{1}{19 \times 21} \\ & =\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+\ldots . .+\frac{1}{2}\left(\frac{1}{19}-\frac{1}{21}\right) \\ & =\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2} \times \frac{20}{21}=\frac{10}{21} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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