Understanding how to use the binomial theorem to expand $(x+y)^3$

Question

The binomial theorem states that:

$$(x+y)^n = \sum_{k=0}^n x^ky^{n-k}$$

Expand $(x+y)^3$:

Here we are given $n=3$ and since the binomial starts at $k=0$, it suffices to say that we are expanding from $k=0$ to $k=3$:

$${3 \choose 0} x^0y^3 + {3 \choose 1} x^1y^2 + {3 \choose 2} x^2y^1 + {3 \choose 3} x^3y^0$$

equates to: $y^3+xy^2+x^2y+x^3$

I don't understand how the equation that I used was supposed to give me this: $$y^3+3xy^2+3x^2y+x^3$$

$n$s and $k$s are only in the exponents. I don't understand why and where the $3$ comes from as a coefficient in the two middle terms!? I know the book says that with ${n \choose k}$, $n$ and $k$ can be thought of as binomial coefficients but I cannot seem to make the connection.

What would $(x+y)^4$ look like? Would it have only $4$'s in as coefficients in the middle terms?

Answer 1

The coefficients come from "N choose K," written as $(^n_k)$ - the first component of the binomial theorem's summation. While you wrote out the problem correctly, you forgot to perform this operation.

For $(^3_0)$, there is one way in which we can select zero elements from a set of three. For $(^3_1)$, there are three ways in which we can select one element from a set of three. For $(^3_2)$ and $(^3_3)$, there are three ways and one way, respectively, for selecting K elements from a set of size N.

Expanding $(x + y)^4$ would not result in four's for the coefficients on the middle terms. You would have to perform the operation I described above. Alternatively, you can use Pascal's Triangle: a triangular arrangement of numbers that represents the coefficient of each term in a binomial expansion.

Answer 2

You can use this to calculate coefficients.

Answer 3

$$ \begin{array}{lcccccccccccccccccccc} \text{row } 0 \rightarrow & & & & & & & 1 \\ \text{row } 1 \rightarrow & & & & & & 1 & & 1 \\ \text{row } 2 \rightarrow & & & & & 1 & & 2 & & 1 \\ \text{row } 3 \rightarrow & & & & 1 & & 3 & & 3 & & 1 \\ \text{row } 4 \rightarrow & & & 1 & & 4 & & 6 & & 4 & & 1 \\ \text{row } 5 \rightarrow & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\ \text{row } 6 \rightarrow & 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \\ & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow & & \uparrow \\[4pt] & \dbinom 6 0 & & \dbinom 6 1 & & \dbinom 6 2 & & \dbinom 6 3 & & \dbinom 6 4 & & \dbinom 6 5 & & \dbinom 6 6 \\ {} \end{array} $$

$$ \text{So } (x+y)^6 = x^6 + 6x^5 y + 15 x^4 y^2 + 20 x^3 y^3 + 15 x^2 y^4 + 6 xy^5 + y^6. $$