回復 1# s7908155 的帖子
\(\begin{align}
& \frac{1}{a+1}+\frac{1}{{{a}^{2}}+1}+\frac{1}{{{a}^{3}}+1}+\cdots +\frac{1}{{{a}^{100}}+1} \\
& =\left( \frac{1}{a+1}+\frac{1}{{{a}^{100}}+1} \right)+\left( \frac{1}{{{a}^{2}}+1}+\frac{1}{{{a}^{99}}+1} \right)+\cdots +\left( \frac{1}{{{a}^{50}}+1}+\frac{1}{{{a}^{51}}+1} \right) \\
& =50 \\
& \\
& \frac{{{a}^{3}}}{a+1}+\frac{{{a}^{6}}}{{{a}^{2}}+1}+\frac{{{a}^{9}}}{{{a}^{3}}+1}+\cdots +\frac{{{a}^{300}}}{{{a}^{100}}+1} \\
& =\frac{{{a}^{3}}+1}{a+1}+\frac{{{a}^{6}}+1}{{{a}^{2}}+1}+\frac{{{a}^{9}}+1}{{{a}^{3}}+1}+\cdots +\frac{{{a}^{300}}+1}{{{a}^{100}}+1}-50 \\
& =\left( {{a}^{2}}-a+1 \right)+\left( {{a}^{4}}-{{a}^{2}}+1 \right)+\left( {{a}^{6}}-{{a}^{3}}+1 \right)+\cdots +\left( {{a}^{200}}-{{a}^{100}}+1 \right)-50 \\
& =\left( {{a}^{102}}+{{a}^{104}}+{{a}^{106}}+\cdots +{{a}^{200}} \right)-\left( a+{{a}^{3}}+{{a}^{5}}+\cdots +{{a}^{99}} \right)+100-50 \\
& =50 \\
\end{align}\)
