Question

Tinh :P= 1/2 +1/2^2 + ... + 1/ 2^99

ai lam duoc minh tich cho

 

Answer 1

\(2P=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

\(2P=1+\frac{1}{2}+...+\frac{1}{2^{98}}\)

\(2P-P=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

\(P=1-\frac{1}{2^{99}}\)

Answer 2

\(P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(2P=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

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

\(2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(P=2P-P=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

\(P=1-\frac{1}{2^{99}}\)

\(P=\frac{2^{99}-1}{2^{99}}\)