\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(2B=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
\(2B-B=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{3^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\right)\)
\(B=1+\frac{3}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(2B=2+\frac{3}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}-\frac{100}{2^{99}}\)
\(2B-B=\left(2+\frac{3}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}-\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\right)\)
\(B=2+\frac{3}{2}+\frac{1}{2^2}-\frac{100}{2^{99}}-1-\frac{3}{2^3}-\frac{1}{2^{99}}+\frac{100}{2^{100}}\)
\(B=2+\frac{3}{2}+\frac{1}{4}-\frac{200}{2^{100}}-1-\frac{3}{8}-\frac{2}{2^{100}}+\frac{100}{2^{100}}\)
\(B=\frac{19}{8}-\frac{102}{2^{100}}=\frac{19}{8}-\frac{51}{2^{99}}\)
