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Math Help - Algebra

  1. #1
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    Algebra

    I am doing proof by induction for my discrete math class by I'm horrible at algebra and I can't figure this last part out. Thanks in advance for the advice.

    I need to make: (10^k) -1 + (9)(10^k)

    look like this: 10^(k+1) -1

    I also need help with this one:

    I need to make: 1 - 1/(k+1) + 1/(k+1)(k+2) (1 is numerator, rest is Denom)

    look like this: 1 - 1/(k+2)
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  2. #2
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    Quote Originally Posted by jzellt View Post
    ...

    I need to make: (10^k) -1 + (9)(10^k)

    look like this: 10^(k+1) -1

    ...
    10^k - 1 +9 \cdot 10^k = 10^k - 1 +(10 -1) \cdot 10^k = 10^k - 1 +10 \cdot 10^k - 10^k=  -1 + 10 \cdot 10^k = 10^{k+1}-1
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  3. #3
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    Thank you! Any advice for the second problem?
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  4. #4
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    Quote Originally Posted by jzellt View Post
    ...

    I also need help with this one:

    I need to make: 1 - 1/(k+1) + 1/(k+1)(k+2) (1 is numerator, rest is Denom)

    look like this: 1 - 1/(k+2)
    1-\frac1{k+1}+\frac1{(k+1)(k+2)}= 1-\frac{k+2}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = 1- \frac{k+2-1}{(k+1)(k+2)}

    Nowyou can cancel (k+1). You should add to the final result that k \ne -1
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  5. #5
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    I got up until the final step you have. After I find the common denominator of (k+1)(k+2), I add up the number and get (k+2) + 1. How did you get (k+2) - 1?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    I got up until the final step you have. After I find the common denominator of (k+1)(k+2), I add up the number and get (k+2) + 1. How did you get (k+2) - 1?
    the minus sign in front of the \frac {k + 2}{(k + 1)(k + 2)} is what yielded the -1

    note: - \frac {k + 2}{(k + 1)(k + 2)} + \frac 1{(k + 1)(k + 2)} = \frac {1 - (k + 2)}{(k + 1)(k + 2)} = - \frac {(k + 2) - 1}{(k + 1)(k + 2)}

    got it?

    can you continue?
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