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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
2 years, 10 months ago

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

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

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

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

\( = \dfrac{1}{d} \times \left( \dfrac{1}{n} - \dfrac{1}{n+d} \right) \) Siddhartha Srivastava · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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@Rachit Mahendra But i still don't understand the exponents of n part. Can you please explain futher. Peter Bishop · 2 years, 10 months ago

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@Peter Bishop 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 · 2 years, 10 months ago

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@Rachit Mahendra Awesome!!!!!!!!!!!!!!! Thank you very much. Peter Bishop · 2 years, 10 months ago

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Woah, cool bro! Finn Hulse · 2 years, 10 months ago

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wow...good formula for rememberd Ben Habeahan · 2 years, 10 months ago

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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 · 2 years, 10 months ago

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This method is use to solve many problems and is called telescoping method I think. Anurag Pandey · 5 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 · 2 years, 10 months ago

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I just learnt A+B=0 so, A=-B wa0w. R I Newton yet? Ayush Banerjee · 2 years, 10 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 · 2 years, 10 months ago

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Awesome method! From where did you get this?? Nashita Rahman · 2 years, 10 months ago

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@Nashita Rahman From a tutor Peter Bishop · 2 years, 10 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 · 2 years, 10 months ago

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