Question

tính tổng:

a)S= 1+ \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3}\)+.....+\(\frac{1}{2^{2018}}\)

mk bít là khó lắm nhưng pn nào học giỏi giúp mk với nhé! tick cho nha

Answer 1

\(2S=2+1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2017}}\)

\(2S-S=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2018}}\right)\)

\(\Rightarrow S=2-\frac{1}{2^{2018}}+1-1+\frac{1}{2}-\frac{1}{2}+.....+\frac{1}{2^{2017}}-\frac{1}{2^{2017}}=2-\frac{1}{2^{2018}}\)\(=\frac{2^{2019}-1}{2^{2018}}\)

Answer 2

bảo bình chứng tỏ S <1 nhé