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What is the value of \(\frac34+\frac5{36}+\frac7{144}+........+\frac{17}{5184}+\frac{19}{8100} ?\)(a) 0.95 (b) 1 (c) 0.99 (d) 0.98 |
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Answer» (c) 0.99 \(\frac34+\frac5{36}+\frac7{144}+........+\frac{17}{5184}+\frac{19}{8100} \) = \(\frac{3}{1^2.2^2}\) + \(\frac{5}{2^2.3^2}\) + \(\frac{7}{4^2.3^2}\) + ........ + \(\frac{17}{8^2.9^2}\) + \(\frac{19}{9^2.10^2}\) = \(\big(1-\frac1{2^2}\big) + \big(\frac1{2^2}-\frac1{3^2}\big) + \big(\frac1{3^2}-\frac1{4^2}\big) + ...... +\big(\frac{1}{8^2}-\frac1{9^2}\big) + \big(\frac{1}{9^2}-\frac1{10^2}\big)\) = 1 - \(\frac1{10^2}\) = 1 - \(\frac1{100} = \frac{99}{100}\) = 0.99. |
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