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Math Help - geometric seies

  1. #1
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    geometric seies

    Use a/k + a/k(1-1/k)+.....a/k(1-1/k)^n+....=a to show that
    3/4 +3/16 + 3/64 + ....+ 3/4^n....= 1
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  2. #2
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    Quote Originally Posted by bigb View Post
    Use a/k + a/k(1-1/k)+.....a/k(1-1/k)^n+....=a to show that
    3/4 +3/16 + 3/64 + ....+ 3/4^n....= 1
    Given formula: \frac{a}{k} \left[1 + \left(1 - \frac{1}{k}\right) + \, .... \, + \left(1 - \frac{1}{k}\right)^n + \, .... \right] = a.

    Required to find: \frac{3}{4} \left(1 + \frac{1}{4} + \, .... \, + \frac{1}{4^{n-1}} + \frac{1}{4^n} + \, .... \right).

    So substitute a = 1 and k = 4/3 into the given formula ......
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    Quote Originally Posted by mr fantastic View Post
    Given formula: \frac{a}{k} \left[1 + \left(1 - \frac{1}{k}\right) + \, .... \, + \left(1 - \frac{1}{k}\right)^n + \, .... \right] = a.

    Required to find: \frac{3}{4} \left(1 + \frac{1}{4} + \, .... \, + \frac{1}{4^{n-1}} + \frac{1}{4^n} + \, .... \right).

    So substitute a = 1 and k = 4/3 into the given formula ......
    i get the series to be 15/16 and it is suppose to be 1
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  4. #4
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    Quote Originally Posted by bigb View Post
    i get the series to be 15/16 and it is suppose to be 1
    Obviously the answer is 1 ..... because a = 1. I gave you that in my first reply.

    Unless you explain how you got your answer I have no idea what mistakes you've made.
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    Quote Originally Posted by mr fantastic View Post
    Obviously the answer is 1 ..... because a = 1. I gave you that in my first reply.

    Unless you explain how you got your answer I have no idea what mistakes you've made.
    I got it. It was a dumb mistake on my part
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