Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(5A-A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)\)

\(4A=1-\frac{1}{5^{100}}\)

\(A=\frac{1-\frac{1}{5^{100}}}{4}\)

\(A=\frac{1}{4}-\frac{1}{5^{100}}:4\)

\(A=\frac{1}{4}-\frac{1}{5^{100}.4}\)

=> \(V=4.5^{100}.\left(\frac{1}{4}-\frac{1}{5^{100}.4}\right)+1\)

\(V=\left(4.5^{100}.\frac{1}{4}-4.5^{100}.\frac{1}{5^{100}.4}\right)+1\)

\(V=\left(5^{100}-1\right)+1\)

\(V=5^{100}\)

\(2.A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)

=> 2.A - A = \(\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)

=> A = \(\left(2+\frac{3}{2^2}-1-\frac{100}{2^{100}}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)\)

A = \(1+\frac{3}{2^2}-\frac{100}{2^{100}}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}=\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)+\frac{2}{2^2}-\frac{100}{2^{100}}\)

Tính B = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

2.B = \(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\) => 2.B - B = \(1+\frac{1}{2}-\frac{1}{2^{99}}\)=> B = \(\frac{3}{2}-\frac{1}{2^{99}}\)

Vậy A = \(\frac{3}{2}-\frac{1}{2^{99}}+\frac{2}{2^2}-\frac{100}{2^{100}}=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}=2=\frac{2^{101}-102}{2^{100}}\)

\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)

=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)=     0

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)+x=2+\frac{1}{5}\)

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}+x=\frac{11}{5}\)

\(\frac{1}{100}+x=\frac{11}{5}\)

\(x=\frac{11}{5}-\frac{1}{100}=\frac{219}{100}\)

đáp án bằng 2 ok

 X=219/100 đó nha!

chúc may mắn!!!