35. If $S=\frac{1}{1+1^{2}+1^{4}}+\frac{2}{1+2^{2}+2^{4}}+\frac{3}{1+3^{2}+3^{4}}+\ldots \ldots \ldots+\frac{200}{1+200^{2}+200^{4}}$, find the value of $80402 \times S$.
35. Answer: 40200
Note that $\frac{k}{1+k^{2}+k^{4}}=\frac{1}{2}\left[\frac{1}{k(k-1)+1}-\frac{1}{k(k+1)+1}\right]$ for all integers $k$.
Hence the required sum can be found by the method of difference
$$
\begin{array}{l}
=\frac{1}{2}\left(\frac{1}{1 \times 0+1}-\frac{1}{1 \times 2+1}+\frac{1}{2 \times 1+1}-\frac{1}{2 \times 3+1}+\frac{1}{3 \times 2+1}-\frac{1}{3 \times 4+1}+\ldots \ldots+\right. \\
=\frac{1}{2}\left(1-\frac{1}{40201}\right) \\
=\frac{20100}{40201}
\end{array}
$$

Hence $80402 \times S=40200$.