(a-3b)^5 expand the binomial please

Question

anonymous · Answer

are you familiar with the binomial theorem?

rainbow_dash · Answer

No.

anonymous · Answer

basicaly it is thus: $$(a+b)^{n} = a^{n}+ na^{n-1}b+\frac{ n(n-1) }{ 2! }a^{n-2}b^{2}+\frac{ n(n-1)(n-2) }{ 3! }a^{n-3}b^{3}+...+b^{n}$$

rainbow_dash · Answer

good god.

anonymous · Answer

lol its faster than using the FOIL rule to multiply out each pair of brackets

rainbow_dash · Answer

:( i'll try to use..thanks lol

anonymous · Answer

post your answer so i can see if you used it ryt. Do you have a calculator with the ! function?

rainbow_dash · Answer

No i do not.

anonymous · Answer

basicaly all that that ! function symbolises is the product of all whole numbers including the one shown. eg 8! = 1*2*3*4*5*6*7*8 = 40320

rainbow_dash · Answer

okay it's ultiple chocie so her'es the one i chose:

$$a^5+5a^4b+10a^3b^2+10a^2b^3+5b^4+b^5$$

anonymous · Answer

yup that right.

anonymous · Answer

these types of quesions can be seen easily becaus of the patern that emerges. as you see the exponents go (5,0)(4,1)(3,2)(2,3)(1,4)(0,5)

rainbow_dash · Answer

cool thanks! in your formula you gave me, when you get tot he part with the factorials, is the part next to the factorial multiplied by the $$\frac{ n(n-1) }{ 2! }$$