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Thread: Mathematical Induction

  1. #1
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    Mathematical Induction

    Hi there,

    I'm totally dumbfounded by this question. Could anyone give me a basis on where to start so I can at least try to work it out myself?

    Use mathematical induction to show that:

    $\displaystyle S(n) = 3*2^{n-1} -2$
    is the solution for the reccurence relation:
    $\displaystyle T(n)=2T(n-1)+2$ for $\displaystyle n > 1$ and $\displaystyle T(1) = 1$
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  2. #2
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    To prove $\displaystyle T(n)=S(n)$ for all positive integers $\displaystyle n$ all you have to do is show

    (i) that $\displaystyle T(1)=S(1)$, and
    (ii) that if $\displaystyle T(k-1)=S(k-1)$ for some $\displaystyle k>1$ then also $\displaystyle T(k)=S(k)$ for that very same $\displaystyle k$.

    Here we go.

    Let $\displaystyle S(n)=3\times2^{n-1}-2$.

    Initial step: $\displaystyle S(1)=3\times2^0-2=3-2=1$, so $\displaystyle T(1)=S(1)$.

    Inductive step: Assume that $\displaystyle T(k-1)=S(k-1)$ for some $\displaystyle k>1$.

    Then $\displaystyle T(k)=2T(k-1)+2=2S(k-1)+2=2(3\times2^{k-2}-2)+2=3\times2^{k-1}-2=S(k)$.

    Thus, by the principle of induction, $\displaystyle T(n)=S(n)$ for all positive integers $\displaystyle n$.
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  3. #3
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    Thanks, that helped a lot. I have a follow up question here.

    If 1 is added to the recurrence relation such that:

    $\displaystyle T(n) = 2T(n-1) + 3$ for $\displaystyle n > 1$ and $\displaystyle T(1) = 0$

    What is the new equation for S(n)?
    Prove it by induction.
    Am I basically reversing the steps in the first answer to come up with a new equation for S(n)? This is what I've come up with:

    $\displaystyle T(k) = 2T(k-1)+3=2S(k-1)+3=3(3*2^{k-2}-2)+3=3x2^{k-1}-3=S(k)$
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