1. ## Binomial expansion

what is the third term in the expansion of $(a-3b)^4$
Also the middle term of the expansion $($sinx $+2)^4$

2. Originally Posted by ~berserk
what is the third term in the expansion of $(a-3b)^4$
Also the middle term of the expansion $($sinx $+2)^4$

The binomial theorem states that

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

Hint the first term is when n=0.

So what would the third term be.

I hope this helps.
Good luck.

3. I just have one more question. For example 5th term in $(3a-b)^6$ i get $(15)(3)^2a^2b^4$ but it is $-b$ does that make a difference or is the answer $135a^2b^4$

4. Originally Posted by ~berserk
I just have one more question. For example 5th term in $(3a-b)^6$ i get $(15)(3)^2a^2b^4$ but it is $-b$ does that make a difference or is the answer $135a^2b^4$
In general yes, in this case no. Here is why.

Consider

$(a-b)^5=[a+(-b)]^5$
$=a^5(-b)^0+5a^4(-b)^1+10a^3(-b)^2+10a^2(-b)^3+5(a)^1(-b)^4+a^0(-b)^5=$

$a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5$

I hope this helps

5. ## Expand x.sin2x upto the term containing x7

Any Ideas, i'm sooooooo stuck

6. Originally Posted by Squeeka7
Any Ideas, i'm sooooooo stuck
You wrote this

$x\sin(2x)=2x\sin(x)\cos(x)$

Did you mean $(x+\sin(2x))^7$

7. Erm no, to give a bit of back ground the question is part of an Analytical methods assignment, and is written "Expand x.sin2x upto the term containing x7"

8. Originally Posted by Squeeka7
Erm no, to give a bit of back ground the question is part of an Analytical methods assignment, and is written "Expand x.sin2x upto the term containing x7"
Well then I assume it is a Maclaurin series you want

$x\sin(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n+2}}{(2n+1)!}$

So what you want is this

$\sum_{n=0}^{3}\frac{(-1)^n(2x)^{2n+2}}{(2n+1)!}$

But there will be no 7 term, a 6 and 8 term though

P.S. This is hardly an Algebra/Pre-algebra question.