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avatar+845 

please explain about the part of multiplying fractions with x^2 as a denominator

 Nov 18, 2018
 #1
avatar+6185 
+3

I think this is supposed to be a demonstration of the binomial theorem.

 

\((a+b)^n = \sum \limits_{k=0}^n~\dbinom{n}{k}a^k b^{n-k}\)

 

In this case

 

\(a = 2\\ b= \dfrac{3}{x^2}\)

 

\(\left(2 + \dfrac{3}{x^2}\right)^4 = \sum \limits_{k=0}^4~\dbinom{4}{k}2^k\left(\dfrac{3}{x^2}\right)^{4-k}\)

 

you can plug in k=4, 0 to get a-e

 Nov 18, 2018
 #2
avatar+845 
+1

yes i know how to do that but i am not sure on the part of after simplifying everything which leaves me with an interger to multiply k/x^2. this is where i struggle. thank you for helping

YEEEEEET  Nov 18, 2018
 #3
avatar+6185 
+3

I don't really understand what you are saying but for example take k=3

 

\(\dbinom{4}{3}2^3 \left(\dfrac{3}{x^2}\right)^{4-3} = \\ 4\cdot 8 \cdot \dfrac{3}{x^2} = \\ 32\dfrac{3}{x^2} = \dfrac{96}{x^2}\\ b=96\)

Rom  Nov 18, 2018
 #4
avatar+845 
+1

this is exactly what i was confused by!

thank you very much!

YEEEEEET  Nov 18, 2018
 #5
avatar
0

\((a + b)^n =\sum_{k=0}^{n} \begin{pmatrix} n\\ k \end{pmatrix}a^\left ( n -k \right )b^k\)

The powers of a and b are reversed in Rom's LaTeX.

 Nov 19, 2018
 #6
avatar+6185 
+1

the two are identical

Rom  Nov 19, 2018

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