(1)/(3)+(1)/(3^(2))+(1)/(3^(3))+...+(1)/(3^(99))+(1)/(3^(100))
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1. It takes me 20 minutes to walk to school every day.
2. Jess spends 30 minutes removing her makeup everyday.
3. It takes her 20 minutes to wash her dog every week.
4. I spend 30 minutes going to school.
5. It takes us 30 minutes to review our lesson before class
6. It took him lots of time to do this experiment.
7. It took Anna three days to visit VN
8. It took her 2 days to tidy her room.
9. It take them half and hour to go to school.
10. It took Nam 5 days to visit Danang.
11. he spent 30 minutes writting this letter.
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Question 1 : I like music and art
Question 2 : no , she isn't
no , she is a dentist
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1 Yes, they are
2 She likes fried potatoes
3 I would like mangoes
4 Yes, she does
Đặt `A= 1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100)`
`3A= 3. (1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100))`
`3A= 1 + 1/3 + 1/(3^2) + ... + 1/(3^98) + 1/(3^99)`
`3A - A = (1 + 1/3 + 1/(3^2)+... + 1/(3^98) + 1/(3^99)) - (1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100))`
`2A = 1 - 1/(3^100)`
`A = (1 - 1/(3^100))/2`
Vậy: `1/3 + 1/(3^2) + 1/(3^3) + ... + 1/(3^99) + 1/(3^100) = (1-1/(3^100))/2`
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{99}}\) + \(\dfrac{1}{3^{100}}\)
3A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{3^{98}}\) + \(\dfrac{1}{3^{99}}\)
3A - A = (1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + ...+\(\dfrac{1}{3^{98}}\) + \(\dfrac{1}{3^{99}}\)) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{99}}\)+\(\dfrac{1}{3^{100}}\))
A.(3 - 1) = 1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{98}}\)+ \(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\) - ...- \(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3^{100}}\)
A x 2 = (1 - \(\dfrac{1}{3^{100}}\)) + (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^{98}}\) - \(\dfrac{1}{3^{98}}\)) + (\(\dfrac{1}{3^{99}}\) - \(\dfrac{1}{3^{99}}\))
A x 2 = 1 - \(\dfrac{1}{3^{100}}\) + 0 + 0 + ..+ 0
A x 2 = 1 - \(\dfrac{1}{3^{100}}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{100}}\)