1. ## 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?

2. Originally Posted by afeasfaerw23231233
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:

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

$\Rightarrow Ak + Bk + A = 1$

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

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

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

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

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

4. Originally Posted by afeasfaerw23231233
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?
Originally Posted by Reckoner
This is just a partial fraction decomposition. Assuming $k$ is not constant:

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

$\Rightarrow Ak + Bk + A = 1$

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

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

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

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

Let $k=0$

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

Now let $k=-1$

giving us

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