Câu 5. (1 điểm): a) Tính A = $\frac{1}{-2}$.$\frac{1}{3}$ + $\frac{1}{-3}$.$\frac{1}{4}$ + ... + $\frac{1}{-9}$.$\frac{1}{10}$. b) Chứng minh rằng: A = $\frac{1}{3}$ + $(\frac{1}{3})^{2}$ + $(\frac{1}{3})^{3}$ + ... + $(\frac{1}{3})^{99}$ < $\frac{1}{2}$

a)

`A = 1/(-2) . 1/3 + 1/(-3) . 1/4 + ... + 1/(-9) . 1/10`

`= - (1/(2.3) + 1/(3.4) + ... + 1/(9.10))`

`= - (1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)`

`= - (1/2 - 1/10)`

`= - 4/10`

`= -2/5`

b)

`A = 1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^99`

`=> 3A = 1 + 1/3 + (1/3)^2 + ... + (1/3)^98 `

`=> 3A - A = [1 + 1/3 + (1/3)^2 + ... + (1/3)^98] - [1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^99]`

`=> 2A = 1 - (1/3)^99`

`=> A = (1 - (1/3)^99)/2 < 1/2` (đpcm)