The sum of 10 terms of the series

Question:

The sum of 10 terms of the series

$\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots . .$ is :

  1. 1

  2. $\frac{120}{121}$

  3. $\frac{99}{100}$

  4. $\frac{143}{144}$


Correct Option: , 2

Solution:

$S=\frac{2^{2}-1^{2}}{1^{2} \times 2^{2}}+\frac{3^{2}-2^{2}}{2^{2} \times 3^{2}}+\frac{4^{2}-3^{2}}{3^{2} \times 4^{2}}+\ldots$

$=\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]+\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]+\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]+\ldots+\left[\frac{1}{10^{2}}-\frac{1}{11^{2}}\right]$

$=1-\frac{1}{121}$

$=\frac{120}{121}$

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