Question

[Vũ Vân Anh]

a ) A = 1/4 + 1/4^2 +1/4^3 +.........+ 1/4^100 + 1/3.4^100

b) B = 1/3 - 1/3^2 + 1/3^3 - 1/3^4 +.........+ 1/ 3^99

Answer 1

[Nguyễn Thanh Hằng]

a)Ta có :

\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)

\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)

\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)

\(3A=1-\dfrac{1}{4^{100}}\)

\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)

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