The sum of the first 99 terms of the series $$\frac{3}{4} + \frac{5}{{16}} + \frac{7}{{144}} + \frac{9}{{400}} + .....$$

Explanation:-

$$\eqalign{ & {\text{Given expression,}} \cr & {\text{ = }}\frac{{4 - 1}}{{4 \times 1}} + \frac{{9 - 4}}{{9 \times 4}} + \frac{{16 - 9}}{{16 \times 9}} + ......... \cr & = \left( {1 - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{9}} \right) + \left( {\frac{1}{9} - \frac{1}{{16}}} \right) + ........ \cr & = \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + .... \cr & \therefore \,99{\text{th term of the series}} \cr & {\text{ = }}\left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right) \cr & \therefore \,\,\,{\text{Given expression,}} \cr & \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) + .... + \left( {\frac{1}{{{{98}^2}}} - \frac{1}{{{{99}^2}}}} \right) + \left( {\frac{1}{{{{99}^2}}} - \frac{1}{{{{100}^2}}}} \right) \cr & = \left( {1 - \frac{1}{{{{100}^2}}}} \right) \cr & = \left( {1 - \frac{1}{{10000}}} \right) \cr & = \frac{{9999}}{{10000}} \cr}$$