A) \[\frac{1}{100}\]
B) \[\frac{99}{100}\]
C) \[\frac{101}{100}\]
D) 1
Correct Answer: B
Solution :
Expression \[=\frac{3}{{{1}^{2}}{{.2}^{2}}}+\frac{5}{{{2}^{2}}{{.3}^{2}}}+\frac{7}{{{3}^{2}}{{.4}^{2}}}+....+\frac{17}{{{8}^{2}}{{.9}^{2}}}+\frac{19}{{{9}^{2}}{{.10}^{2}}}\] \[=\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)+\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)+\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[...+\left( \frac{1}{{{8}^{2}}}-\frac{1}{{{9}^{2}}} \right)+\left( \frac{1}{{{9}^{2}}}-\frac{1}{{{10}^{2}}} \right)\] \[=\frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+......\] \[\frac{1}{{{8}^{2}}}-\frac{1}{{{9}^{2}}}+\frac{1}{{{9}^{2}}}-\frac{1}{{{10}^{2}}}\] \[=1-\frac{1}{{{10}^{2}}}\] \[=1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}\]You need to login to perform this action.
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