To solve (a+b)^{n}, Pascal's triangle may help you:
1 n=0
1 1 n=1
1 2 1 n=2
1 3 3 1 n=3
1 4 6 4 1 n=4
1 5 10 10 5 1 n=5
1 6 15 20 15 6 1 n=6
1 7 21 35 35 21 7 1 n=7
1 8 28 56 70 56 28 8 1 n=8
I've shown the first 9 lines, however, you can continue writing lines indefinitely. Every number is the sum of upper two. For example in n=3 line, 1 = 0 + 1, 2 = 1 + 1, 1 = 1 + 0. In n=4 line 1 = 0 + 1, 3 = 1 + 2, 3 = 2 + 1, 1 = 1 + 0 and so on.
Numbers in these lines coefficients for expansion of (a+b)^{n}.
(a+b)^{n} can be expanded to Ca^{n}+Ca^{n-1}b+Ca^{n-2}b^{2}+...+Ca^{2}b^{n-2}+Cab^{n-1}+Cb^{n} where C are the coefficients from line n.
For example:
(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}
To expand (2+3x)^{5} we must take the coefficients from n=5 line =>
(2+3x)^{5}=2^{5}+5*2^{4}*3x+10*2^{3}*(3x)^{2}+10*2^{2}*(3x)^{3}+5*2*(3x)^{4}+(3x)^{5} = 32+270x+720x^{2}+1080x^{3}+810x^{4}+243x^{5}
If I calculated correctly, than from this you can see that the coefficient of x cubed is 1080. I think that expanding the binomial this way is much easier than multiplying.
I hope that helps.
P.S. If you see the triangle with a straight angle, than look http://en.wikipedia.org/wiki/Pascal_triangle" .
