Find the eighth term of (3a+b)^9

Homeschooled so any help would be appreciated thanks.

Question

Find the eighth term of (3a+b)^9

Homeschooled so any help would be appreciated thanks.

highschoolmom2010 · Answer

this is just 
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anonymous · Answer

easier with the binomial theorem

whpalmer4 · Answer

yeah, the 8th term of that... suppose the problem was to find the 97th term of $$ (3a+b)^{126}$$? :-)

the answer is to use the binomial theorem...

whpalmer4 · Answer

the binomial theorem allows you to calculate any term directly, without having to expand that monster (which by the way is:$$6561 a^8+17496 a^7 b+20412 a^6 b^2+13608 a^5 b^3+5670 a^4 b^4$$$$+1512 a^3 b^5+252 a^2 b^6+XXX+b^8$$

whpalmer4 · Answer

(I've blocked out the 8th term so that we can have the fun of finding it)

whpalmer4 · Answer

do you know anything about the binomial theorem?

anonymous · Answer

OHHH so it is kinda like using pascal's triangle? I haven't learned much about it yet but I will look into it, thanks for the help!

whpalmer4 · Answer

yes, pascal's triangle gives you the binomial coefficients...there's also a way to compute them directly if you don't want to build the triangle.

anonymous · Answer

I'm a bit confused how wouuld I start?

whpalmer4 · Answer

once you know the binomial coefficient for the term of interest, you just multiply it by the appropriate powers of x and y.

say we are doing $$(x+y)^n$$(in our problem, $x = 3a,~ y = b,~n = 8$)

the binomial coefficient is written $\binom{n}{k}$ and can be found by 
$$\binom{n}{k} = \frac{n!}{(n-k)!k!}$$ where $n! = n*(n-1)*(n-2)...2*1$

$k$ is a variable that tells us which term we are finding.  It goes from 0 to $n$.

The "appropriate powers of $x$ and $y$" are

$$x^{n-k}y^{k}$$ 

So, let's do an easy example:

$$(x+2)^3$$We could multiply this out by hand easily enough:
$$(x+2)^3 = (x+2)(x+2)(x+2) = (x^4+4x+4)(x+2) = x^3+6x^2+12x+8$$
Let's see how we would find the 3rd term with the binomial theorem.

$$n = 3, k = 3-1=2, x = x, y = 2$$
$$\binom{3}{2} = \frac{3!}{(3-2)!2!} = \frac{3*2*1}{1*2*1} = 3$$
Our term is $$\binom{n}{k}x^{n-k}y^{k} = \binom{3}{2}x^{3-2}y^{2} = 3x^1y^2 $$But remember that $y = 2$ so that becomes $$3x^1(2)^2 = 12x$$Sure enough, $12x$ is the 3rd term when we consult our hand-multiplied version.

anonymous · Answer

Thank you! Just got the answer!

whpalmer4 · Answer

In your problem, we have $$x = 3a,~y=b, n = 8, k=8-1 =7$$
So the 8th term is
$$\binom{8}{7} x^7y^1 = \frac{8!}{(8-7)!7!} (3a)^1(b)^7 = 8*(3a)b^7 = 24ab^7 $$