ABC26GN1086 · Simplification
Subject: General Aptitude · Chapter: Simplification · Exam: 2004 · Marks: · Difficulty:
The value of $\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\frac{9}{4^{2} \cdot 5^{2}}+\frac{11}{5^{2} \cdot 6^{2}}+$ $\frac{13}{6^{2} \cdot 7^{2}}+\frac{15}{7^{2} \cdot 8^{2}}+\frac{17}{8^{2} \cdot 9^{2}}+\frac{19}{9^{2} \cdot 10^{2}}$ is
(a)$\frac{1}{100}$
(b)$\frac{99}{100}$
(c)1
(d)$\frac{101}{100}$
Answer
Explanation
Given $\exp .=\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$ $$\begin{gathered} +\left(\frac{1}{4^{2}}-\frac{1}{5^{2}}\right)+\ldots . .+\left(\frac{1}{9^{2}}-\frac{1}{10^{2}}\right) \\ =\left(\frac{1}{1^{2}}-\frac{1}{10^{2}}\right)=\left(1-\frac{1}{100}\right)=\frac{99}{100} . \end{gathered}$$
Explanation as extracted from the printed page; notation may be imperfect.
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