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Math Help - identity

  1. #1
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    identity

    p20 q36
    question:
    \frac 1 {k(k+1)} \equiv \frac A k + \frac B {k+1}
    where A and B are constants. find the value of A and B.

    A(k+1) +Bk=1
    then how to do?
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p20 q36
    question:
    \frac 1 {k(k+1)} \equiv \frac A k + \frac B {k+1}
    where A and B are constants. find the value of A and B.

    A(k+1) +Bk=1
    then how to do?
    This is just a partial fraction decomposition. Assuming k is not constant:

    A(k+1) + Bk = 1

    \Rightarrow Ak + Bk + A = 1

    \Rightarrow k(A + B) + A = 0k + 1

    \Rightarrow ({\color{red}A + B})k + {\color{blue}A} = {\color{red}0}k + {\color{blue}1}

    \Rightarrow\left\{<br />
\begin{array}{rcl}<br />
{\color{red}A + B} & {\color{red}=} & {\color{red}0}\\<br />
{\color{blue}A} & {\color{blue}=} & {\color{blue}1}<br />
\end{array}<br />
\right.
    Last edited by Reckoner; June 7th 2008 at 10:06 AM. Reason: Added color for emphasis and clarity
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Or just do \frac{1}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.

    Hence A=1 & B=-1
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p20 q36
    question:
    \frac 1 {k(k+1)} \equiv \frac A k + \frac B {k+1}
    where A and B are constants. find the value of A and B.

    A(k+1) +Bk=1
    then how to do?
    Quote Originally Posted by Reckoner View Post
    This is just a partial fraction decomposition. Assuming k is not constant:

    A(k+1) + Bk = 1

    \Rightarrow Ak + Bk + A = 1

    \Rightarrow k(A + B) + A = 0k + 1

    \Rightarrow ({\color{red}A + B})k + {\color{blue}A} = {\color{red}0}k + {\color{blue}1}

    \Rightarrow\left\{<br />
\begin{array}{rcl}<br />
{\color{red}A + B} & {\color{red}=} & {\color{red}0}\\<br />
{\color{blue}A} & {\color{blue}=} & {\color{blue}1}<br />
\end{array}<br />
\right.
    Reckoner's way is completely valid, but personally I think it is easier when applicable (i.e. when your factors have real roots) to do this

    1=A(k+1)+Bk

    Let k=0

    giving us 1=A(0+1)+B(0)\Rightarrow{A=1}

    Now let k=-1

    giving us

    1=A(1+-1)+B(-1)\Rightarrow{B=-1}
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