The value of frac{3}{1^{2} cdot 2^{2}}+frac{5 | vedclass

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(B) The given series is $\sum_{n=1}^{9} \frac{2n+1}{n^{2}(n+1)^{2}}$.We know that $\frac{2n+1}{n^{2}(n+1)^{2}} = \frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}

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$\frac{99}{100}$

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(B) The given series is $\sum_{n=1}^{9} \frac{2n+1}{n^{2}(n+1)^{2}}$.We know that $\frac{2n+1}{n^{2}(n+1)^{2}} = \frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}} = \frac{1}{n^{2}} - \frac{1}{(n+1)^{2}}$.Substituting this into the series:$= (\frac{1}{1^{2}} - \frac{1}{2^{2}}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{3^{2}} - \frac{1}{4^{2}}) + \dots + (\frac{1}{9^{2}} - \frac{1}{10^{2}})$.This is a telescoping series where all intermediate terms cancel out.$= \frac{1}{1^{2}} - \frac{1}{10^{2}} = 1 - \frac{1}{100} = \frac{99}{100}$.

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

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