Question

bài 1: Tính

a, \(A=1+3+3^2+3^3+....3^n\)

b, \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+....+\frac{1}{10^{100}}\)

 

Answer 1

a.

\(A=1+3+3^2+3^3+...+3^n\)

\(3A=3+3^2+3^3+3^4+...+3^{n+1}\)

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

\(2A=3^{n+1}-1\)

\(A=\frac{3^{n+1}-1}{2}\)

b.

\(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\)

\(10B=10+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^{98}}+\frac{1}{10^{99}}\)

\(10B-B=\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\right)-\left(10+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^{98}}+\frac{1}{10^{99}}\right)\)

\(9B=\frac{1}{10^{100}}-10\)

\(B=\frac{\frac{1}{10^{100}}-10}{9}\)