Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ A=\frac{1}{2}\left(\frac{1}{3^{20}} +1\right)\\ b.\ B=\frac{1}{124}\left(\frac{3}{5^{100}} +75\right)\\ c.\ C=5^{100} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ A=\frac{1}{3} +\frac{1}{3^{2}} +\frac{1}{3^{3}} +...+\frac{1}{3^{20}}\\ \Leftrightarrow \frac{1}{3} A=\frac{1}{3^{2}} +\frac{1}{3^{3}} +...+\frac{1}{3^{21}}\\ \Leftrightarrow \frac{1}{3} A-A=\left(\frac{1}{3^{2}} +\frac{1}{3^{3}} +...+\frac{1}{3^{21}}\right) -\left(\frac{1}{3} +\frac{1}{3^{2}} +...+\frac{1}{3^{20}}\right)\\ \Leftrightarrow -\frac{2}{3} A=\frac{1}{3^{21}} -\frac{1}{3} \Leftrightarrow A=\frac{1}{2}\left(\frac{1}{3^{20}} +1\right)\\ b.\ B=\frac{3}{5} +\frac{3}{5^{4}} +...+\frac{3}{5^{100}}\\ \Leftrightarrow \frac{1}{5^{3}} B=\frac{3}{5^{4}} +\frac{3}{5^{7}} +...+\frac{3}{5^{103}}\\ \Leftrightarrow \frac{1}{5^{3}} B-B=\left(\frac{3}{5^{4}} +\frac{3}{5^{7}} +...+\frac{3}{5^{103}}\right) -\left(\frac{3}{5} +\frac{3}{5^{4}} +...+\frac{3}{5^{100}}\right)\\ \Leftrightarrow \frac{-124}{125} B=\frac{3}{5^{103}} -\frac{3}{5} \Leftrightarrow B=\frac{1}{124}\left(\frac{3}{5^{100}} +75\right)\\ c.\ C=4.5^{100} .\left(\frac{1}{5} +\frac{1}{5^{2}} +...+\frac{1}{5^{100}}\right) +1\\ \Leftrightarrow C=4.\left( 5^{99} +5^{98} +...+1\right) +1=4D+1\\ Ta\ có:\ 5D=5^{100} +5^{99} +...+5\\ \Leftrightarrow 5D-D=\left( 5^{100} +5^{99} +...+5\right) -\left( 5^{99} +5^{98} +...+1\right)\\ \Leftrightarrow 4D=5^{100} -1\\ \Leftrightarrow C=5^{100} -1+1=5^{100} \end{array}$
