Question

\(B=\frac{1}{7}+\frac{1}{7^{2}}+\frac{1}{7^{3}}+...+\frac{1}{7^{100}}\)

Answer 1

\(B=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\)

\(7B=7\cdot\left(\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\right)\)

\(7B=1+\dfrac{1}{7}+...+\dfrac{1}{7^{99}}\)

\(7B-B=\left(1+\dfrac{1}{7}+....+\dfrac{1}{7^{99}}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\right)\)

\(6B=1-\dfrac{1}{7^{100}}\)

\(B=\dfrac{1-\dfrac{1}{7^{100}}}{6}\)

\(B=\dfrac{1}{6}-\dfrac{1}{6\cdot7^{100}}\)