Question

Hi, I can't figure out what I'm supposed to do on this problem. Could someone help? Thanks.

Find the coefficient of a of the term in the expansion of the binomial.

Binomial
(x^2+3)^12

Term
ax^8

Answer 1

the General term(r+1)
= C(12,r)(x^2)^(12-r)(3^r)
= C(12,r)(x^(24-2r)(3^r)

so ax^8 = C(12,r)(x^(24-2r)(3^r)

then 24-2r = 8
r = 8

It must be the 9th term and it must be
C(12,8)x^8(3^8)
= 495(6561)x^8

so a = 3247695

Answer 2

It wants the number in from of the x^8 term
but in (p+q)^12 = p^12 + 12 p^11 q + 66 p^10 q^2 +220p^9q^3+ 495 p^8q^4 ......
we want the term with p^4 q^8 (which is the same as the term for p^8q^4) because it is x^2 and not x
If you either use a binomial expansion table or Pascal's triangle or the formula for binomial coefficients the coefficient of the (12,4) term is 495
C(12,4) = C(12,8) = 12!/(4!*8!) = 12*11*10*9 /4*3*2 = 11*5*9 =495

Answer 3

Whoops, sorry, forgot the 3^8

Answer 4

2r³+250