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C2 binomial expansion question

Hey guys this is a binomial expansion question I'm having trouble understanding the method , any help would be greatly appreciated 😎 " Find the coefficient of x^11 in the binomial expansion of (9/2x - 2x^2/3)^13

(edited 6 years ago)

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Original post by Hardy jacks

Hey guys this is a binomial expansion question I'm having trouble understanding the method , any help would be greatly appreciated 😎 " Find the coefficient of x^11 in the binomial expansion of (9/2x - 2x^2/3)^13

maybe u can help..

Study Forum Helper

Original post by Hardy jacks

Hey guys this is a binomial expansion question I'm having trouble understanding the method , any help would be greatly appreciated " Find the coefficient of x^11 in the binomial expansion of (9/2x - 2x^2/3)^13

Clearly you want

x^{11}

So in the expansion, there is a term whereby, as far only the x terms are concerned,

(\frac{1}{x})^{\text{something}}\cdot (x^2)^{13-\text{something}}=x^{11}

So if I say that the "something" exponent is

a

then

-a+26-2a=11

This leaves

a=5

It's difficult to explain, but basically you have to find two values which add to make 13 that can be the powers of both

\frac{9}{2x}

and

-\frac{2x^2}{3}

, such that you would get

\frac{(x^2)^a}{x^b}=x^{11}

.

In this case,

a=8, b=5

its hard to explain
try to work out what power u need to raise the a and b in (a+b)^13
when its (1/x -x^2)^13

then sub 9/2x for 1/x
and 2x^2/3 for b

you want to work out when the powers of x cancel out to make 11 basically

OP

Original post by RDKGames

Clearly you want

x^{11}

So in the expansion, there is a term whereby, as far only the x terms are concerned,

(\frac{1}{x})^{\text{something}}\cdot (x^2)^{13-\text{something}}=x^{11}

So if I say that the "something" exponent is

a

then

-a+26-2a=11

This leaves

a=5

Thank you so much , this makes perfect sense how do you get 8? 😊

Study Forum Helper

Original post by Hardy jacks

Thank you so much , this makes perfect sense how do you get 8? 😊

That's the 13-a

OP

Original post by RDKGames

That's the 13-a

Of course! Thank you for the last minute help

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