Question

[|๖ۣۜRPK-ACE(♔⋆şɦเฑเςɦเ⋆♔)]

a) Cho A=1 phần 2+(1 phần 2)^2+(1 phần 2)^3+...+(1 phần 2)^99

CMR:A<1

b) Cho B=1 phần 3 + (1 phần 3^2) + (1 phần 3^3) + ... + (1 phần 3^99)

CMR:B<1 phần 2

Answer 1

[Huỳnh Quang Sang]

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

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

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

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

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