Question

Tính tổng

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

\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

Answer 1

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

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

\(2B-B=1-\frac{1}{2^{100}}\)

\(B=1-\frac{1}{2^{100}}\)

\( C=\frac{1}{2}-\frac{1}{2^2}+.................+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(2 C=1-\frac{1}{2}+......................+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)

\(2 C+C=1-\frac{1}{2^{100}}\)

\(C=\left(1-\frac{1}{2^{100}}\right):3\)