A nice way to factorize FRACTIONS!

I recently learned that ...

1n(n+d)\frac{1}{n(n+d)} = 1d(1n1n+d)\frac{1}{d}(\frac{1}{n}-\frac{1}{n+d})

For example,

135=13(3+2)=12(1315)\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 135+157+179+1911\frac{1}{3*5}+\frac{1}{5*7}+\frac{1}{7*9}+\frac{1}{9*11}

=12(1315)+12(1517)+12(1719)+12(19111)=\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

12[(1315)+(1517)+(1719)+(19111)]\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.

12[13111]\frac{1}{2}[\frac{1}{3}-\frac{1}{11}]

which is 433\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, 7 months ago

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1 vote

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

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

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

=1d×(n+dn(n+d)nn(n+d)) = \dfrac{1}{d} \times \left(\dfrac{n+d}{n(n+d)} - \dfrac{n}{n(n+d)} \right)

=1d×(1n1n+d) = \dfrac{1}{d} \times \left( \dfrac{1}{n} - \dfrac{1}{n+d} \right)

Siddhartha Srivastava - 5 years, 7 months ago

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By using Partial Fractions,let - 1n(n+d) \frac{1}{n(n+d)} = An+Bn+d \frac{A}{n} + \frac{B}{n+d}
Thus, we need to find the coefficients A and B. Therefore,by taking LCM,we get -
1n(n+d) \frac{1}{n(n+d)} = A(n+d)+Bnn(n+d) \frac{A(n+d) + B \cdot n}{n(n+d)} . Or, 1=n(A+B)+Ad 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 n0 n^0 on both sides),

n0:Ad=1,or,A=1d n^0 : A \cdot d =1, or , A = \frac{1}{d}

(Comparing the terms which contains n1 n^1 on both sides),

n1:A+B=0,or,B=A, n^1 : A + B =0 , or, B = -A ,

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

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

1n(n+d) \frac{1}{n(n+d)} = 1d(1n1n+d) \frac{1}{d} ( \frac{1}{n} - \frac{1}{n+d})

Rachit Mahendra - 5 years, 7 months ago

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

Peter Bishop - 5 years, 7 months ago

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

Peter Bishop - 5 years, 7 months ago

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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)=anxn+an1xn1++a1x+a0P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0 and Q(x)=bnxn+bn1xn1++b1x+b0Q(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots +b_1x+b_0 such that P(x)=Q(x)P(x)=Q(x), then ai=bia_i=b_i for all i=0ni=0\to n.

Daniel Liu - 5 years, 7 months ago

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

Bogdan Simeonov - 5 years, 7 months ago

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wow...good formula for rememberd

Ben Habeahan - 5 years, 7 months ago

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Woah, cool bro!

Finn Hulse - 5 years, 7 months ago

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

Peter Bishop - 5 years, 7 months ago

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Awesome method! From where did you get this??

Nashita Rahman - 5 years, 7 months ago

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From a tutor

Peter Bishop - 5 years, 7 months ago

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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 - 5 years, 7 months ago

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

Ayush Banerjee - 5 years, 7 months ago

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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 - 5 years, 7 months ago

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

Anurag Pandey - 3 years, 2 months ago

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