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Math Help - summation question

  1. #1
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    summation question

    Hey guys I can't solve the second part of this following question, i.e. 'hence, ...' onwards.

    Prove by induction that, for all positive integers n, [(end value=n) (summation notation, sigma) (starting value: r=1)] (r+1)2^r = n[2^(n+1)].

    Hence, deduce that [(end value=2n) (summation notation, sigma) (starting value: r=n+1)] [(r+1)2^r + 2^(n+1)] = n[2^(2n+2)].

    Sorry it's so messy but I don't have the programs for 'typing math'. Thanks in advance if anybody could help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by margaritas View Post
    Hey guys I can't solve the second part of this following question, i.e. 'hence, ...' onwards.

    Prove by induction that, for all positive integers n, [(end value=n) (summation notation, sigma) (starting value: r=1)] (r+1)2^r = n[2^(n+1)].

    Hence, deduce that [(end value=2n) (summation notation, sigma) (starting value: r=n+1)] [(r+1)2^r + 2^(n+1)] = n[2^(2n+2)].

    Sorry it's so messy but I don't have the programs for 'typing math'. Thanks in advance if anybody could help!
    You have:

    <br />
\sum_1^{n} (r+1)2^r = n[2^{n+1}] \ \ \ \ ...(1)<br />

    Then:

    <br />
\sum_1^{2n} (r+1)2^r - \sum_1^{n-1} (r+1)2^r = <br />
\sum_{n}^{2n} (r+1)2^r <br />

    So just use (1) to replace the two sums on the left hand side with their
    respective totals to get the result.

    RonL
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  3. #3
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    Thanks but I still don't get it!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    You have:

    <br />
\sum_1^{n} (r+1)2^r = n[2^{n+1}] \ \ \ \ ...(1)<br />

    Then:

    <br />
\sum_1^{2n} (r+1)2^r - \sum_1^{n-1} (r+1)2^r = <br />
\sum_{n}^{2n} (r+1)2^r <br />

    So just use (1) to replace the two sums on the left hand side with their
    respective totals to get the result.

    RonL
    Quote Originally Posted by margaritas View Post
    Thanks but I still don't get it!
    You split the sum from 1 to 2n into two parts one from 1 to n-1 and the
    other fron n to 2n. The formula you already have:

    <br />
\sum_1^{n} (r+1)2^r = n[2^{n+1}] \ \ \ \ ...(1)<br />

    tells you that:

    <br />
\sum_1^{2n} (r+1)2^r = 2n[2^{2n+1}]<br />

    and that:

    <br />
\sum_1^{n-1} (r+1)2^r = (n-1) 2^{n}<br />

    So:

    <br />
\sum_{n}^{2n} (r+1)2^r  = \sum_1^{2n} (r+1)2^r - \sum_1^{n-1} (r+1)2^r = 2n[2^{2n+1}] - (n-1) 2^{n}<br />


    RonL
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  5. #5
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    Hello, margaritas!

    I've got the first part . . .


    Prove by induction that, for all positive integers n:
    . . \sum_{r = 1}^n (r+1)\cdot2^r \:= \:n\cdot2^{n+1}

    Verify that S(1) is true.
    . . S(1) \:=\:(1 + 1)\cdot2^1 \:=\:1\cdot2^2\quad\Rightarrow\quad 4 = 4 . . . true!

    Assume that S(k) is true:
    . . \sum_{r=1}^k(r + 1)\cdot2^r \:=\:k\cdot2^{k+1}

    Add (k+2)\cdot2^{k+1} to both sides:
    . . \sum_{r=1}^k(r+1)\cdot2^r + (k+2)\cdot2^{k+1} \;=\;k\cdot2^{k+1} + (k+2)\cdot2^{k+1}


    The left side is: . \sum^{k+1}_{r=1} (r+1)\cdot2^r . . . the left side of S(k+1).


    The right side is: . \underbrace{k\cdot2^{k+1} + k\cdot2^{k+1}} +\,2^{k+2}
    . . . . . . . . . . . . . . = \;2k\cdot2^{k+1} + 2^{k+2}

    . . . . . . . . . . . . . . = \;k\cdot2^{k+2} + 2^{k+2}

    . . . . . . . . . . . . . . = \;(k+1)\cdot2^{k+2} . . . the right side of S(k+1)


    The inductive proof is complete.

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