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Thread: Induction with summing

  1. #1
    Super Member Deadstar's Avatar
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    Induction with summing

    Now i usually can do induction problems but i cant figure out what to do when theres summing involved. Can anyone help with the following question?

    Prove by induction:
    $\displaystyle \sum_{k=1}^n (-1)^2 k^2 = \frac{1}{2}(-1)^nn(n+1)$

    It works for n = 1 so we have our inductive phase...
    Now assuming n is true...
    How do we prove this for n + 1?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Deadstar View Post
    Now i usually can do induction problems but i cant figure out what to do when theres summing involved. Can anyone help with the following question?

    Prove by induction:
    $\displaystyle \sum_{k=1}^n (-1)^2 k^2 = \frac{1}{2}(-1)^nn(n+1)$

    It works for n = 1 so we have our inductive phase...
    Now assuming n is true...
    How do we prove this for n + 1?
    you're missing something. (-1)^2 = 1, it would make no sense to write it. so there should probably be an n in the power or something. so you left that off, maybe other things
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  3. #3
    Super Member PaulRS's Avatar
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    It is $\displaystyle
    \sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
    {2}
    $, right?

    Suppose $\displaystyle
    \sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
    {2}
    $ holds for n (1)

    We'll now show that: $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{\left( {n + 2} \right) \cdot \left( {n + 1} \right)}}
    {2}
    $ is also true

    Indeed: $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \sum\limits_{k = 1}^n {\left( { - 1} \right)^k \cdot k^2 } + \left( { - 1} \right)^{n + 1} \cdot \left( {n + 1} \right)^2
    $

    But, by (1) we have: $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \cdot \frac{{n \cdot \left( {n + 1} \right)}}
    {2} + \left( { - 1} \right)^{n + 1} \cdot \left( {n + 1} \right)^2
    $

    So: $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^n \left( {n + 1} \right) \cdot \left[ {\frac{n}
    {2} - \left( {n + 1} \right)} \right] = \left( { - 1} \right)^n \left( {n + 1} \right) \cdot \left[ {\frac{{ - n - 2}}
    {2}} \right]

    $

    $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\left( { - 1} \right)^k \cdot k^2 } = \left( { - 1} \right)^{n + 1} \cdot \frac{{\left( {n + 1} \right) \cdot \left( {n + 2} \right)}}
    {2}
    $
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  4. #4
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    Hello, Deadstar!

    Prove by induction: .$\displaystyle \sum_{k=1}^n (-1)^2 k^2 \:= \:\frac{1}{2}(\text{-}1)^nn(n+1)$
    $\displaystyle S(n)\!:\;\;-1^2 + 2^3 - 3^2 + 4^2 + \cdots + (\text{-}1)^nn^2 \;=\;\frac{1}{2}(\text{-}1)^nn(n+1)$


    Verify $\displaystyle S(1)\!:\;\;-1^2 \:=\:\frac{1}{2}(\text{-}1)(1)(2)\quad\Rightarrow\quad -1 \:=\:-1$ . . . True!


    Assume $\displaystyle S(k)\!:\;\;-1^2 + 2^2 - 3^2 + 4^2 + \cdots + (\text{-}1)^kk^2 \;=\;\frac{1}{2}(\text{-}1)^kk(k+1)$



    Add $\displaystyle (\text{-}1)^{k+1}(k+1)^2$ to both sides:

    . . $\displaystyle -1^2 + 2^3 - 3^2 + 4^2 + \cdots + (\text{-}1)^{k+1}(k+1)^2 \;=\;\frac{1}{2}(\text{-}1)^kk(k+1) + (\text{-}1)^{k+1}(k+1)^2$ . [1]


    Factor the right side: .$\displaystyle \frac{1}{2}(\text{-}1)^k(k+1)\,\bigg[k - 2(k+1)\bigg] \;=\;\frac{1}{2}(\text{-}1)^k(k+1)[-k-2]$

    . . $\displaystyle =\;\frac{1}{2}(\text{-}1)^k(k+1)(\text{-}1)[k+2] \;=\;\frac{1}{2}(\text{-}1)^{k+1}(k+1)(k+2) $


    Hence, [1] becomes: .$\displaystyle -1^2 + 2^3 - 3^2 + 4^3 + \cdots + (\text{-}1)^{k+1}(k+1)^2 \;=\;\frac{1}{2}(\text{-}1)^{k+1}(k+1)(k+2)$

    . . This is $\displaystyle S(k+1)$ . . . The inductive proof is complete.

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  5. #5
    Super Member Deadstar's Avatar
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    Thanks everybody, the (-1)^2 was a typo, was meant to be, as you all guessed, (-1)^k. Cheers!
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