Question

[Trần Việt Hoàng]

Tính nhanh:

1/3+1/3²+1/3³+...+1/3²⁰¹⁸

1+5+5²+5³+...+5²⁰¹⁸

Answer 1

[Trần Thanh Phương]

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2017}}\)

\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)\)

\(2A=1-\frac{1}{3^{2018}}\)

\(A=\frac{1-\frac{1}{3^{2018}}}{2}\)

Answer 2

[Nguyễn Phương Uyên]

đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)\)

\(2A=1-\frac{1}{3^{2018}}\)

\(A=\frac{1-\frac{1}{3^{2018}}}{2}\)

đặt \(B=1+5+5^2+...+5^{2018}\)

\(5B=5+5^2+5^3+...+5^{2019}\)

\(5B-B=\left(5+5^2+5^3+...+5^{2019}\right)-\left(1+5+5^2+...+5^{2018}\right)\)

\(4B=5^{2019}-1\)

\(B=\frac{5^{2019}-1}{4}\)