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Math Help - induction

  1. #1
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    induction

    I have the cartesian product of this:

    1 - (1/(k^2)) from k = 2 to n.

    I picked random values of n like n = 2, 3, 4.

    With n = 2, I have (1 - (1/4)).
    With n = 3, I have (1 - (1/4)) * (1 - (1/9)).
    With n = 4, I have (1 - (1/4)) * (1 - (1/9)) * (1 - (1/16)).
    etc.

    How do I go about finding a formula for this for all n?

    I know that 1 - (1/(k^2)) = ((k + 1)(k - 1))/k^2.

    Thanks in advance for any help.
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  2. #2
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    Hello, PvtBillPilgrim!

    I have the cartesian product of this: . 1 - \frac{1}{k^2}\;\text{ for }k \,= \,2\text{ to }n.

    How do I go about finding a formula for this?
    You should have examined those cases more closely . . .


    \begin{array}{cccc}n = 2: & 1 - \frac{1}{2^2} & = & \frac{3}{4} \\ \\<br /> <br />
n = 3: & \left(\frac{3}{4}\right)\left(1-\frac{1}{3^2}\right) & = & \frac{4}{6} \\ \\<br /> <br />
n = 4: & \left(\frac{4}{6}\right)\left(1-\frac{1}{4^2}\right) & = & \frac{5}{8} \\ \\<br /> <br />
n=5: & \left(\frac{5}{8}\right)\left(1-\frac{1}{5^2}\right) & = & \frac{6}{10} \end{array}


    It appears that the general form is: . P_n \;=\;\frac{n+1}{2n}

    The next task is to prove it . . .

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