ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

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

mà 1/2^99>0 nên B<1 (đpcm)

C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99

=> C = 1/3^99 = 1/(3^99) 

=> C < 1/2 (đpcm) 

2A=2^101-2^100+2^98+...+2^3-2^2

3A = 2A + A

3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )

A = (2^101-2) :3

B tăng tự 

a, \(A=\frac{1}{2}+\left[\frac{1}{2}\right]^2+\left[\frac{1}{2}\right]^3+...+\left[\frac{1}{2}\right]^{99}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

\(2A-A=\left[1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right]-\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right]\)

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

Do đó A < 1

b, \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(3B-B=\left[1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]-\left[1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right]\)

\(2B=1-\frac{1}{3^{99}}\)

\(B=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

ôi zời ghi từng bài thôi @_@

nhiều quá duyetj đi