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?

#Algebra #AlgebraicIdentities #AlgebraicManipulation #KeyTechniques #AlgebraicExpansion

Easy Math Editor

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Comments

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

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

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

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

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

Siddhartha Srivastava - 7 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, $\frac{1}{n(n+d)}$ = $\frac{\frac{1}{d}}{n} + \frac{\frac{-1}{d}}{n+d}$

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

Rachit Mahendra - 7 years, 3 months ago

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

Peter Bishop - 7 years, 3 months ago

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

Peter Bishop - 7 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 $i=0\to n$ .

Daniel Liu - 7 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.

Bogdan Simeonov - 7 years, 3 months ago

wow...good formula for rememberd

Ben Habeahan - 7 years, 3 months ago

Woah, cool bro!

Finn Hulse - 7 years, 3 months ago

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

Peter Bishop - 7 years, 3 months ago

Awesome method! From where did you get this??

Nashita Rahman - 7 years, 3 months ago

From a tutor

Peter Bishop - 7 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

Vignesh Subramanian - 7 years, 3 months ago

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

Ayush Banerjee - 7 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

Nupur Ambani - 7 years, 3 months ago

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

Anurag Pandey - 4 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$