Question

Tính 1/3 +1/3^2+1/3^3+.....+1/3^50

Answer 1

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{50}}\)

=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{49}}\)

=>\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{49}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{50}}\right)\)

=>2A=\(1-\frac{1}{3^{50}}\)

=>A=\(\frac{1-\frac{1}{3^{50}}}{2}\)

\(=>A=\frac{1}{2}-\frac{1}{\frac{3^{50}}{2}}=\frac{1}{2}-1.\frac{2}{3^{50}}=\frac{1}{2}-\frac{2}{3^{50}}=\frac{3^{50}-4}{2.3^{50}}\)

Vậy..................