Question

CMR:

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

Answer 1

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

\(\Rightarrow2y=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2y-y=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow y=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<1\)

Answer 2

ta có : 2y=\(\frac{1}{2}+\frac{1}{^{2^2}}+...+\frac{1}{2^{99}}\)

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

y=0,5=>y<1

Answer 3

2y=1/2+1/2^2+...+1/2^99

2y-y=1/2-1/2^100

y=0,5 =>y<1