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Math Help - Simple induction algebra

  1. #1
    Member Jones's Avatar
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    Simple induction algebra

    Hi,

    I were reviewing some induction and saw this:

    prove that for all integers n:
    \sum_{k=1}^n \frac{k^2}{2^k} = 6-\frac{n^2+4n+6}{2^n}

    And the solution looks like this:

    \sum_{k=1}^{n+1} \frac{k^2}{2^k}6-~\frac{n^2+4n+6}{2^n} + \frac{(n+1)^2}{2^{n+1}} \longrightarrow

    \frac{2n^2+8r+12-r^2-2r-1}{2^{r+1}}=...

    Whats going on here? why did they subtract r^2+2r+1
    And the reason they multiplied the expression by 2 where to get the same denominator, but the denominator in the new expression is still 2^{n+1} why?
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  2. #2
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    Smile

    hi
    \sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}, proceed like this (from where they got the r ? )
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  3. #3
    Member Jones's Avatar
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    Quote Originally Posted by Raoh View Post
    hi
    \sum_{k=1}^{n+1}\frac{k^{2}}{2^{k}}=6-\frac{n^2+4n+6}{2^{n}}+\frac{(n+1)^2}{2^{n+1}}, proceed like this (from where they got the r ? )
    Sorry, the 'r's are supposed to be n's
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  4. #4
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    try to expand what i gave you,you should end up with something like this,
    \frac{6(2^n+1)-(n+1)^2-4(n+1)-6}{2^{n+1} }
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  5. #5
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    Quote Originally Posted by Raoh View Post
    try to expand what i gave you,you should end up with something like this,
    \frac{6(2^n+1)-(n+1)^2-4(n+1)-6}{2^{n+1} }
    Hi,

    Hm, to get the whole expression over 2^{n+1} you multiplied by 2 ?
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  6. #6
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    i just added the term \left [\frac{(n+1)^2}{2^{n+1}}  \right ].
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  7. #7
    Member Jones's Avatar
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    Quote Originally Posted by Raoh View Post
    i just added the term \left [\frac{(n+1)^2}{2^{n+1}}  \right ].
    But they have different denominators?
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  8. #8
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    yes,but you have 6-\frac{n^2+4n+6}{2^{n}} = \frac{6\times2^n-n^2-4n-6}{2^n}=\frac{2(6\times2^n-n^2-4n-6)}{2^{n+1}}.
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  9. #9
    Member Jones's Avatar
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    Quote Originally Posted by Raoh View Post
    yes,but you have 6-\frac{n^2+4n+6}{2^{n}} = \frac{6\times2^n-n^2-4n-6}{2^n}=\frac{2(6\times2^n-n^2-4n-6)}{2^{n+1}}.
    neat!

    Thanks!
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