Question

Cm: A<1/3

A=1/4+1/4^2+1/4^3+...+1/4^99

Giúp mình vs mai mình thi r

Answer 1

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

4A=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{4^{98}}\)

4A-A=(1+\(\dfrac{1}{4}\)+..+\(\dfrac{1}{4^{98}}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{4^{99}}\))

3A=1-\(\dfrac{1}{4^{99}}\)

A=

Answer 2

A=\(\dfrac{1-\dfrac{1}{4^{99}}}{3}\)