Question

TÍNH  A= 1/3+1/3^2+1/3^3+...+1/3^99

SO SÁNH A VỚI 1/2

Answer 1

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

\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.........+\frac{1}{3^{100}}-\left(\frac{1}{3}+\frac{1}{3^2}+.......+\frac{1}{3^{99}}\right)=1+\frac{1}{3}\)

\(\Rightarrow2A=1+\frac{1}{3}\Rightarrow A=\left(1+\frac{1}{3}\right):2\)

Answer 2

=>3A=1/3^2+1/3^3+1/3^4+...+1/3^100

=>3A-A=(1/3^2+1/3^3+1/3^4+...+1/3^100) - (1/3+1/3^2+1/3^3+...+1/3^99)

=>2A=1/3^100-1/3

=>A=(\(\frac{1}{3^{100}}\)- \(\frac{1}{3}\)):2

Li ke mình nha!