need to expand Binomials: (a + 3b)^3, etc.

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Hi i'm just fishing up some homework and I'm wating help on 1 and want to know if I got the right answer on another two.

Expanding binomials
first for (a+3b)^3 is it a^3+9a^2b+27ab^2+27b^3?

ok now i need help on expansion in the third term

(x+2y)^5 is itt 60x^3y^2 the answer

the last my dad helped me with but I don't think we got it right

expansion of the third term again
(a+3b)7 what we got is 189a^5b^2

i'm so lost on expanding pretty sure i got the rest right
 
Re: First post so kinda lost need to expand Binomial

Ace-of-Decks said:
Hi i'm just fishing up some homework and I'm wating help on 1 and want to know if I got the right answer on another two.

Expanding binomials
first for (a+3b)^3 is it a^3+9a^2b+27ab^2+27b^3?correct

ok now i need help on expansion in the third term

(x+2y)^5 is it 40x^3y^2 the answer

the last my dad helped me with but I don't think we got it right

expansion of the third term again
(a+3b)^7 what we got is 189a^5b^2correct

i'm so lost on expanding pretty sure i got the rest right
Click to expand...
 
\(\displaystyle \L
\left( {a + b} \right)^n = \sum\limits_{k = 0}^n {n \choose k} a^k b^{n-k
}\)
 
You know there's an old trick to finding the coefficients of a binomial

expansion. The exponents are easy. Add to one, subtract from the other.


Let's take \(\displaystyle (a+b)^{6}\)

As you know, we start with \(\displaystyle a^{6}\) as the first term.

The second term has as its coefficent the exponent of the first term. So, we now have:

\(\displaystyle a^{6}+6a^{5}b\). Follow so far?.

Now........multiply the coefficient of the second term by the exponent of its corresponding a, then divide by the exponent of b plus 1.

Here we go.....(6*5)/2=15. That's the coefficient of the third term. See?.

\(\displaystyle a^{6}+6a^{5}b+15a^{4}b^{2}\)

Now do it again....(15*4)/3=20

\(\displaystyle a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{3}\)

Now, we have gotten to the part where the exponents are the same.

It is now symmetric and repeats. So we have in its entirety:

\(\displaystyle a^{6}+6a^{5}b+15a^{4}b^{2}+20a^{3}b^{3}+15a^{2}b^{4}+6ab^{5}+b^{6}\)

If the exponent of the binomial expansion were odd, say 5, then when you get to the point where the a exponent is 1 more than the b exponent , then you repeat.

For instance, \(\displaystyle (a+b)^{5}\) would be:

\(\displaystyle a^{5}+5a^{4}b+10a^{3}b^{2}+\text{...now repeat...}+10a^{2}b^{3}+5ab^{4}+b^{5}\)

Clear as mud?.
 
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