Question

Thu gọn các tổng sau:

a. A=8.5¹⁰⁰.(\(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)) +1

b. B=\(\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)

Answer 1

a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)

\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)

\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)

\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)

Answer 2

b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)

\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)

Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)

\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)

\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)