Could someone give me a hand at this one?
What is the coefficient of x[5th power] in the expansion of (x+3)[8th power]?
Possible solutions:
56
336
504
1512
Hello, Gretchen,
as you know
$\displaystyle (a+b)^n=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\e nd{array}\right) a^{n-k} \cdot b^k$
and
$\displaystyle \left(\begin{array}{c}n\\k\end{array}\right)=\frac {n!}{k! \cdot (n-k)!}$
ou got:
n = 8
k = 3 (then you get x^(8-3) = x^5)
Now plug in the values you know:
$\displaystyle \left(\begin{array}{c}8\\3\end{array}\right)=\frac {8!}{3! \cdot (5)!}= \frac{40320}{6 \cdot 120}=56$
EB
PS.: I'm not Hänsel
For less computation re the binomial coefficients, use the Pascal's Triangle.
(x+3)^8
The x^5 term is in the 4th term of the expansion.
The 4th term, according to the Pascal Triangle, has 56 as the binomial coefficient.
So the 4th term is 56(x^5)(3^3) = 1512x^5
Therefore, the coefficient of the x^5 is 1512. -----------answer.
--------------------
By the nCr method,
The binomial coefficient of the 4th term is
8C3 = (8!) / (3!)(8-3)!
= (8!) / (3!)(5)!
= (8*7*6*) / (3*2*1)
= 56
Then, continue as above.
Hello, Gretchen!
Don't you know the Binomial Expansion?
If you don't, you shouldn't be trying this problem . . .
What is the coefficient of x^5 in the expansion of (x + 3)^8 ?
Possible solutions: . 56 . . 336 . . 504 . . 1512
You should be familiar with the "combination" numbers of the expanation.
C(8,8)x^8 + C(8,7)(x^7)(3) + C(8,6)(x^6)(3^2) + C(8,5)(x^5)(3^3) + C(8,4)(x^4)(3^4) + . . .
. . . . . . . . . . . . . . . . . . . . . . 8!
So we have: . C(8,3)(3^3) .= .----- (27) .= .(56)(27) .= .1512
. . . . . . . . . . . . . . . . . . . . . .5!3!