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Math Help - Mathematic Induction

  1. #1
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    Mathematic Induction

    Prove by induction that∑_(r=1)^n▒(r+3) 3^r= 3/4[3^n(2n + 5) 5]

    I only manage to get up to

    ∑_(r=1)^(k+1)▒(r+3) 3^r= 3/4[3^(k+1)(2k + 7) 5]
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  2. #2
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    Quote Originally Posted by azuki View Post
    Prove by induction that∑_(r=1)^n▒(r+3) 3^r= 3/4[3^n(2n + 5) – 5]

    I only manage to get up to

    ∑_(r=1)^(k+1)▒(r+3) 3^r= 3/4[3^(k+1)(2k + 7) – 5]
    What's that funny symbol supposed to be?

    Is it this:

     \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5] ??
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  3. #3
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     \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]

    For  n = 1

     \displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12

    And:

     \frac{3}{4}[3^n(2n+5)-5] = \frac{3}{4}[3^1(2(1)+5)-5] =\frac{3}{4}[16] = 12

    Assume it's true for  n = k

     \displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5]

    Now for  n = k+1 , we expect that the result is identical to the result for  n = k except with one more addition of  (r+3)3^r added.

    Hence:

     \displaystyle \sum_{r = 1}^{k+1} (r+3)3^r = (k+1+3)3^{k+1}+   \sum_{r = 1}^{k} (r+3)3^r  = (k+4)3^k \times 3  +   \sum_{r = 1}^{k} (r+3)3^r

     \displaystyle =  (k+4)3^k \times 3  +  \frac{3}{4}[3^k(2k+5)-5]

    Now, to complete the proof, you need to manipulate that expression into

     \frac{3}{4}[3^{k+1}(2(k+1)+5)-5]

    Can you do that?
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  4. #4
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    Quote Originally Posted by Mush View Post
    What's that funny symbol supposed to be?

    Is it this:

     \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5] ??
    Yes. How do I make the symbol appear correctly in this forum?
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  5. #5
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    Quote Originally Posted by azuki View Post
    Yes. How do I make the symbol appear correctly in this forum?
    Look at my previous post. If you click on the equations, you will see the code needed to produce them.
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  6. #6
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    Here is the proof, more formally:

     \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]

    For  n = 1

     \displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12

    And:

     \frac{3}{4}[3^n(2n+5)-5] = \frac{3}{4}[3^1(2(1)+5)-5] =\frac{3}{4}[16] = 12

    Base case holds.

    Assume it's true for  n = k

     \displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5]

    Now for  n = k+1 , we expect that the result is identical to the result for  n = k except with one more addition of  (r+3)3^r added.

    Hence:

     \displaystyle \sum_{r = 1}^{k+1} (r+3)3^r = (k+1+3)3^{k+1}+   \sum_{r = 1}^{k} (r+3)3^r  = (k+4)3^k \times 3  +   \sum_{r = 1}^{k} (r+3)3^r

     \displaystyle =  (k+4)3^{k+1} +  \frac{3}{4}[3^k(2k+5)-5]

     \displaystyle =   \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5]+(k+4)3^{k+1}

     \displaystyle =   \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5+\frac{4}{3}(k+4)3^{k+1}]

     \displaystyle =   \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4}{3}(k+4))-5]

     \displaystyle =   \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4k}{3}+\fra  c{16}{3})-5]

     \displaystyle =   \frac{3}{4}[3^{k+1}(\frac{6}{3}k+\frac{21}{3})-5]

     \displaystyle =   \frac{3}{4}[3^{k+1}(2k+7)-5]

     \displaystyle =   \frac{3}{4}[3^{k+1}(2(k+1)+5)-5]

    QED.
    Last edited by Mush; January 21st 2009 at 08:05 AM.
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