ABC26GN1088 · Simplification

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