3

a)Find the coefficient of $x^3y^4$ in $(2x + 5y)^7$. b) Find the coefficient of $x^5$ in $(3x -1)(2x +1)^8$.

I know this has to do with generating functions , but i'm not sure how to start with this problem in order to find the coefficient of part a) and b)

naslundx
  • 9,720
user142743
  • 127
  • 2
    Use the binomial theorem. – Git Gud Apr 19 '14 at 21:11
  • 2
    How to start: forget generating functions and concentrate on binomial expansions. – Did Apr 19 '14 at 21:11
  • To explain a bit more why you may be having trouble: this very well might be about generating functions. But you are given the generating functions here. To do the "extract the coefficient" part, you have to know how to expand your GFs into sums or series. That expansion step is what everyone is showing. – ex0du5 Apr 19 '14 at 21:21

4 Answers4

2

You can (and should) do this with the binomial theorem. We expand $(2x+5y)^7$ this way:

$$\binom{7}{0}(2x)^7(5y)^{0} + \binom{7}{1}(2x)^6(5y)^{1} + \binom{7}{2}(2x)^5(5y)^{4} + \dots + \binom{7}{7}(2x)^0(5y)^{7} $$

Specifically, the term we are interested in is

$$\binom{7}{4}(2x)^3(5y)^4$$

which can be simplified to $175000 \cdot x^3y^4$.

The second problem is done in the same way.

naslundx
  • 9,720
  • @user142743 That doesn't seem right. You should expand only the (2x+1)^8 part, and find the two terms which, when multiplied by (3x-1) contain x^5. – naslundx Apr 19 '14 at 21:54
1

For (b), you have $(3x - 1)(2x + 1)^{8}$. So consider $(2x+1)^{8}$. You are interested in the coefficients of $x^{4}$ and $x^{5}$ in that term. When you multiply by $(3x-1)$ you can form $x^{5}$ by $3x * kx^{4}$, and $x^{5}$ is held constant by multiplying with $-1$. So $(2x + 1)^{8}$ has coefficient of $x^{5}$ as $\binom{8}{5} (2x)^{5}$ and coefficient of $x^{4}$ as $\binom{8}{4}$.

Now consider $3 * \binom{8}{4} x^{4} * x - \binom{8}{5} (2x)^{5}$.

ml0105
  • 14,674
0

A related technique. Recalling the Taylor series in two variables the coefficient of $x^3y^4$ is given by

$$\frac{1}{3!4!} \frac{\partial^3}{\partial x^3}\frac{\partial^4}{\partial y^4}(2x+5y)^7\Big |_{(x,y)=(0,0)} = 175000. $$

Community
  • 1
Mhenni Benghorbal
  • 47,431
0

Write(b) as: \begin{align} [x^5] (3 x - 1) (2 x + 1)^8 &= 3 [x^4] (2 x + 1)^8 - [x^5](2 x + 1)^8 \\ &= 3 \binom{8}{4} \cdot 2^4 - \binom{8}{5} \cdot 2^5 \end{align} Use of the "coefficient of" ($[\cdot]$) operator simplifies writing and manipulating such expressions a lot.

vonbrand
  • 27,812