# A nice way to factorize FRACTIONS!

I recently learned that ...

$\frac{1}{n(n+d)}$ = $\frac{1}{d}(\frac{1}{n}-\frac{1}{n+d})$

For example,

$\frac{1}{3*5}=\frac{1}{3(3+2)}=\frac{1}{2}(\frac{1}{3}-\frac{1}{5})$

Which is useful because we can use this equation in situation like this...

Find the value of $\frac{1}{3*5}+\frac{1}{5*7}+\frac{1}{7*9}+\frac{1}{9*11}$

$=\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{9})+\frac{1}{2}(\frac{1}{9}-\frac{1}{11})$

Take the common factor out

$\frac{1}{2}[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{11})]$

The terms cancels each other and we are left with the first and last terms.

$\frac{1}{2}[\frac{1}{3}-\frac{1}{11}]$

which is $\frac{4}{33}$ This way is MUCH faster than the traditional way

Anyone can prove why the equation i learned is true?

Note by Peter Bishop
5 years, 3 months ago

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$\dfrac{1}{n(n+d)}$

$= \dfrac{1}{d} \times \dfrac{d}{n(n+d)}$

$= \dfrac{1}{d} \times \dfrac{(n+d) - (n)}{n(n+d)}$

$= \dfrac{1}{d} \times \left(\dfrac{n+d}{n(n+d)} - \dfrac{n}{n(n+d)} \right)$

$= \dfrac{1}{d} \times \left( \dfrac{1}{n} - \dfrac{1}{n+d} \right)$

- 5 years, 3 months ago

By using Partial Fractions,let - $\frac{1}{n(n+d)}$ = $\frac{A}{n} + \frac{B}{n+d}$
Thus, we need to find the coefficients A and B. Therefore,by taking LCM,we get -
$\frac{1}{n(n+d)}$ = $\frac{A(n+d) + B \cdot n}{n(n+d)}$. Or, $1= n(A + B) + A \cdot d$
Now, comparing the terms for different exponents of 'n' on both sides, we get-

{ This can be explained by example given below -

If ax + b = 3x +2 , then , a = 3 and b = 2 }

(Comparing the terms which contains $n^0$on both sides),

$n^0 : A \cdot d =1, or , A = \frac{1}{d}$

(Comparing the terms which contains $n^1$ on both sides),

$n^1 : A + B =0 , or, B = -A ,$

Or, $B = \frac{-1}{d}$

Therefore, $\frac{1}{n(n+d)}$ = $\frac{\frac{1}{d}}{n} + \frac{\frac{-1}{d}}{n+d}$

$\frac{1}{n(n+d)}$ = $\frac{1}{d} ( \frac{1}{n} - \frac{1}{n+d})$

- 5 years, 3 months ago

Awesome!!!!!!!!!!!!!!! Thank you very much.

- 5 years, 3 months ago

But i still don't understand the exponents of n part. Can you please explain futher.

- 5 years, 3 months ago

He is using a property called "comparing coefficients" (or at least that is what I call it)

It states the following: For any two polynomials $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$ and $Q(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots +b_1x+b_0$ such that $P(x)=Q(x)$, then $a_i=b_i$ for all $i=0\to n$.

- 5 years, 3 months ago

It's pretty easy.Multiply the RHS by $n(n+d)$ and you will get $\frac {1}{d}.[(n+d)-n]=1$ And that's it.

- 5 years, 3 months ago

wow...good formula for rememberd

- 5 years, 3 months ago

Woah, cool bro!

- 5 years, 3 months ago

Guys, if you like what you learnt, please reshare it so others can see. Thanks for the comments

- 5 years, 3 months ago

Awesome method! From where did you get this??

- 5 years, 3 months ago

From a tutor

- 5 years, 3 months ago

when we can write the pattern or series in the form that difference of two numbers and such that it get cancelled it makes the work easier this is called v n method or elemination method

- 5 years, 3 months ago

I just learnt A+B=0 so, A=-B wa0w. R I Newton yet?

- 5 years, 3 months ago

1/n(n+d) =1/d x d/n(n=d) =1/d x (n+d)-(n)/n(n+d) =1/d x {(n+d)/n(n+d-(n)/n(n+d)} =1/d x (1/n-1-n+d) HENCE PROVED

- 5 years, 3 months ago

This method is use to solve many problems and is called telescoping method I think.

- 2 years, 10 months ago