Binomial Expansion

[Arnold G]

I was wondering if anyone here has a great understanding of this topic because I really dont understand questions like

-find the coefficient of x^6 when (2-x)(3x+1)^9

Help would be awesome thanks

[ctrls]

What part don't you understand? The question itself, or how to solve it?

For the question itself, you have to expand the expression and find out what the coefficient of the x^6 term is - the value it is multiplied by. If it, for example ends up being 54x^6, the coefficient will be 54.

As an example, if you consider a simpler case of expanding (1-x)(x+1)^2 where you have to find the x^2 term, you can expand out the brackets and simplify it as,

In which case the coefficient is -1.

For the question you listed, you could similarly expand the (3x+1)^9 using the binomial theorem and go from there. There is however, a faster way of doing this, which may be what you were asking about in the first place,

Although expanding out the entire expression gives the correct answer, it's largely a waste of time since it requires so many steps. For example, there's not point finding the x^9 term in the expansion, because multiplying it with (2-x) will give something far greater than x^6. In fact, you only need to consider the x^5 and x^6 terms of the expansion, since those are the ones which gives the desired value when you multiply it by (2-x).

This can be done rather simply using the binomeal formula, as it states that for (a+b)^n, the k^th term can be expressed as . In this case a=3x, so you will need to consider the (9-5)th and (9-6)th case, so k=3,4. This gives,

Which gives the 91854 as the required solution. It's hard to explain, but I hope that makes sense.

Edit: Sorry, but the formatting changes if I include spaces, so I've had to take them out. It's a bit hard to read, but just look for the equals signs.

Edit2: Fixed.

Edited September 12, 2013 by ctrls

[Arnold G]

Are you an "expert" in this topic? Because I need a lot of help

[ninale]

you need to expand (3x+1)^9 and see which combinations fit you...

so

(2-x)[(3x)⁹+(⁹C₁)(3x)⁸1¹+(⁹C₂)(3x)⁷1²+(⁹C₃)(3x)⁶1³ +(⁹C₄)(3x)⁵1⁴+...]

so you simply need

2(⁹C₃)(3x)⁶1³+(-x)(⁹C₄)(3x)⁵1⁴

in order to have only powers of 6

[ctrls]

Are you an "expert" in this topic? Because I need a lot of help

Er, not really. I'm just another student who happens to take maths at HL, though I do understand the concepts well enough to solve the problem you noted.

Just saying "I need help" however doesn't really mean anything, you really need to give more details. What exactly don't you understand? What exactly do you need help with?