The sum of the first 99 terms of the series 3/4 + 516 + 7144 + 9400 + .....

Answer: Option C

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