Question

[Hà Phúc Thành]

tính:

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

B=1/2^2+1/2^4+...+1/2^100

xin cảm ơn những ai đã giải bài giúp mình nha

Answer 1

[Nobi Nobita]

a) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=A=1-\frac{1}{2^{99}}\)

b) \(B=\frac{1}{2^2}+\frac{1}{2^4}+...........+\frac{1}{2^{100}}\)

\(\Rightarrow2^2.B=4B=1+\frac{1}{2^2}+......+\frac{1}{2^{98}}\)

\(\Rightarrow4B-B=3B=1-\frac{1}{2^{100}}\)

\(\Rightarrow b=\frac{1-\frac{1}{2^{100}}}{3}\)

Answer 2

[Xyz OLM]

a) Ta có  A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

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

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

=> A = \(1-\frac{1}{2^{99}}\)

b) Ta có B = \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)

=> 2²B = \(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}=4B\)

=> 4B - B = \(\left(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\right)\)

=> 3B = \(1-\frac{1}{2^{100}}\)

=> B = \(\frac{1}{3}-\frac{1}{2^{100}.3}\)