Question

[TH]

So sánh:

1)A= \(\frac{1}{2}\)+\(\frac{1}{2^2}\) + \(\frac{1}{2^3}\)+....+\(\frac{1}{2^{49}}\)+ \(\frac{1}{2^{50}}\)với 1

2) B=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+....+\(\frac{1}{3^{99}}\)+\(\frac{1}{2^{100}}\)với \(\frac{1}{2}\)

3)C= \(\frac{1}{4}\)+\(\frac{1}{4^2}\)+\(\frac{1}{4^3}\)+.....+\(\frac{1}{4^{999}}\)+ \(\frac{1}{4^{1000}}\)với \(\frac{1}{3}\)

Answer 1

[Trần Thanh Phương]

\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)

\(2A-A=1-\frac{1}{2^{50}}\)

\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1

tương tự nha

Answer 2

[Edogawa Conan]

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}< 1\)