Question

tính B=1+1 phần 5+1 phần 5 mũ 2+...+1 phần 5 mũ 2018

Answer 1

B=1+1/5+1/5²+...+1/5²⁰¹⁸
=>5B=5+1+1/5+...+1/5²⁰¹⁷
=>5B-B=5-1/5²⁰¹⁸
=>4B=5-1/5²⁰¹⁸
=>B=(5-1/5²⁰¹⁸)/4

Answer 2

\(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)

\(\Rightarrow5B=5\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)

\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\)

\(\Rightarrow5B-B=\left(5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)

\(\Rightarrow4B=5-\frac{1}{5^{2018}}\)

\(\Rightarrow B=\frac{5-\frac{1}{5^{2018}}}{4}\)

Vậy \(B=\frac{5-\frac{1}{5^{2018}}}{4}\)

Answer 3

\(B=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{2018}}\)

\(\Rightarrow5B=5+1+\frac{1}{5}+.....+\frac{1}{5^{2017}}\)

\(\Rightarrow5B-B=\left(5+1+\frac{1}{5}+....+\frac{1}{5^{2017}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)

\(\Rightarrow4B=5-\frac{1}{5^{2018}}\)    \(\Rightarrow B=\frac{5-\frac{1}{5^{2018}}}{4}\)