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
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
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
Whoops, sorry, forgot the 3^8
2r³+250
To find the coefficient of the term ax^8 in the expansion of the binomial (x^2 + 3)^12, you can use the binomial theorem. The binomial theorem states that the coefficient of the term with binomial coefficients nCr, where n is the exponent of the binomial and r is the power of x in the term.
For the term ax^8, the power of x is 8. Now let's determine the exponent of the binomial:
Step 1: Expand the binomial (x^2 + 3)^12 using the binomial theorem.
(x^2 + 3)^n = nC0(x^2)^n + nC1(x^2)^(n-1)*(3^1) + nC2(x^2)^(n-2)*(3^2) + ... + nCr(x^2)^(n-r)*(3^r) + ... + nCn*(3^r)
In this case, n = 12.
Step 2: Identify the term ax^8 within the expansion.
The term ax^8 can be obtained when:
- nCr = a, where r is the number of x^2 terms, and
- (x^2)^r * (3)^(12-r) = x^8.
In other words, we need to solve for r in the equation (x^2)^r * (3)^(12-r) = x^8.
Step 3: Solve for r.
(x^2)^r * 3^(12-r) = x^8
(x^2^r) * (3^(12)) * (3^(-r)) = x^8
(1^r) * (3^(12-r)) * (3^(-r)) = x^8
3^(12-2r) = x^8
Since we have a power of x equal to 8, we can equate the exponents to solve for r:
12 - 2r = 8
-2r = 8 - 12
-2r = -4
r = 2
Step 4: Find the corresponding binomial coefficient.
The coefficient of the term ax^8 is determined by the binomial coefficient nCr, where n = 12 and r = 2:
nCr = 12C2 = 12! / (2!(12-2)!) = 66
Therefore, the coefficient of ax^8 in the expansion of (x^2 + 3)^12 is 66.