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  1. #1
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    a tree is planted when it has a height of 2 metres. duringthe first year it grows to a height of 3 metres. thereafter the increase in height every year is 4/5 of the increase in height durind the previus year. what isthe maximum height that the tree will reach? please people help me out on this one, its a sequence and series question
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  2. #2
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    So, write it out and ponder it...

    2 + 1 + 4/5 + (4/5)^{2} + (4/5)^{3} + ... = 2 + \left(\frac{1}{1-\frac{4}{5}}\right)

    You may be expected to 1) Recognize the structure of the Geometric Series and 2) Know the formula for the sum.

    Note: The question is a little funny. I don't really like the phrase "maximum height that the tree will reach". I would prefer a limit sort of phrase, maybe something like "What height will the tree approach?" or "What is the minimum height the tree will NOT exceed?" It is a tough concept to get exactly right. Maybe the original question is okay for now.
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  3. #3
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    thanx 4 helping, however. i am still puzzled on how u got the answer
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  4. #4
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    Hello,

    Allow me to answer...

    Quote Originally Posted by TKHunny View Post
    So, write it out and ponder it...

    2 + 1 + 4/5 + (4/5)^{2} + (4/5)^{3} + ... = 2 + \left(\frac{1}{1-\frac{4}{5}}\right)

    You may be expected to 1) Recognize the structure of the Geometric Series and 2) Know the formula for the sum.

    Note: The question is a little funny. I don't really like the phrase "maximum height that the tree will reach". I would prefer a limit sort of phrase, maybe something like "What height will the tree approach?" or "What is the minimum height the tree will NOT exceed?" It is a tough concept to get exactly right. Maybe the original question is okay for now.
    Quote Originally Posted by bongz View Post
    thanx 4 helping, however. i am still puzzled on how u got the answer
    Do you understand the first part ?

    --> h=2+1+(4/5)+(4/5)^2+\dots

    You can see that 2 is the original height. The following part is a geometric series.

    1+q+q^2+q^3+\dots+q^n=\frac{1-q^{n+1}}{1-q}

    Here, n \to \infty
    And |q|<1, which means that \lim_{n \to \infty} q^n=0

    So 1+q+q^2+q^3+\dots \to \infty=\frac{1}{1-q}

    Is it more clear ? :x
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  5. #5
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    Quote Originally Posted by bongz View Post
    thanx 4 helping, however. i am still puzzled on how u got the answer
    1) I didn't get the answer. I left that for you.

    2) I gave you a list of two things you are expected to know. Look them up and start knowing them. Only then will you see how YOU will get such answers.
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