1. ## identity

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

$\displaystyle A(k+1) +Bk=1$
then how to do?

2. Originally Posted by afeasfaerw23231233
p20 q36
question:
$\displaystyle \frac 1 {k(k+1)} \equiv \frac A k + \frac B {k+1}$
where A and B are constants. find the value of A and B.

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

$\displaystyle A(k+1) + Bk = 1$

$\displaystyle \Rightarrow Ak + Bk + A = 1$

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

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

$\displaystyle \Rightarrow\left\{ \begin{array}{rcl} {\color{red}A + B} & {\color{red}=} & {\color{red}0}\\ {\color{blue}A} & {\color{blue}=} & {\color{blue}1} \end{array} \right.$

3. Or just do $\displaystyle \frac{1}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.$

Hence $\displaystyle A=1$ & $\displaystyle B=-1$

4. Originally Posted by afeasfaerw23231233
p20 q36
question:
$\displaystyle \frac 1 {k(k+1)} \equiv \frac A k + \frac B {k+1}$
where A and B are constants. find the value of A and B.

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

$\displaystyle A(k+1) + Bk = 1$

$\displaystyle \Rightarrow Ak + Bk + A = 1$

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

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

$\displaystyle \Rightarrow\left\{ \begin{array}{rcl} {\color{red}A + B} & {\color{red}=} & {\color{red}0}\\ {\color{blue}A} & {\color{blue}=} & {\color{blue}1} \end{array} \right.$
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

$\displaystyle 1=A(k+1)+Bk$

Let $\displaystyle k=0$

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

Now let $\displaystyle k=-1$

giving us

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