Self studying Basic matematics, by Lang.
Stuck in this section. Does someone know what “select x from k factors” mean?
Understanding binomial expansion
binomial theoremcombinatorics
Related Solutions
This is probably the wrong proof for you, but I will post it anyways. (requires calculus)
Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.
If it is an analytic function, then it should follow Taylor's theorem.
Now, if we take the expansion around $x=0$, we get
$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$
Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$
or
$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$
$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$
where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.
(b) Consider the previous result as a description of flipping 5 unfair coins. Even though the coin is unfair, which two outcomes are equally likely?
Your guess is correct in question (b)
In the expression $(2h+t)^5$, $h$ represents a head and $t$ represents a tail. $(2h+t)^5$ is actually a description of five identical unfair coins each has $\frac{2}{3}$ chance getting a head and $\frac{1}{3}$ getting a tail. In the expanded form: $$32h^5 + 80h^4t^1 + 80h^3t^2 + 40h^2t^3 + 10ht^4 + t^5$$ We can see that the probabaility of getting 4 heads and 1 tail or getting 3 heads and 2 tails are both $\frac{80}{32+80+80+40+10+1} = \frac{80}{243}$.
(c) Consider again $(2h+t)^{100}$. Imagine you added every coefficient in the expansion. What would be the total?
It is often useful to simplify the case and find the pattern first. From the (b) part, we can see that the coefficient of $32h^5 + 80h^4t^1 + 80h^3t^2 + 40h^2t^3 + 10ht^4 + t^5$ add up to $243$, which is $3^5$. By following this pattern, we can find out the answer of (c) is $3^{100}$.
You may want to know why the pattern is like this. Here is the justification. Let's consider$(2h+t)^5$ again. When we plug $h=1$ and $t=1$, the equation becomes $$32(1)^5 + 80(1)^4(1)^1 + 80(1)^3(1)^2 + 40(1)^2(1)^3 + 10(1)(1)^4 + (1)^5 = 243$$
It is now clear that we can find the sum of coefficient just by plugging 1 in all the variables. Therefore the answer will be $$(2(1)+1)^{100} = 3^{100}$$
Best Answer
Let's look at the examples $(x + y)^3$ and $(x + y)^3$.
By direct computation, we obtain \begin{align*} (x + y)^3 & = (x + y)(x + y)^2\\ & = (x + y)[(x + y)(x + y)]\\ & = (x + y)[x(x + y) + y(x + y)]\\ & = (x + y)(x^2 + xy + xy + y^2)\\ & = (x + y)(x^2 + 2xy + y^2)\\ & = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)\\ & = x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3\\ & = x^3 + 3x^2y + 3xy^2 + y^3 \end{align*} and \begin{align*} (x + y)^4 & = (x + y)(x + y)^3\\ & = (x + y)(x^3 + 3x^2y + 3xy^2 + y^3)\\ & = x(x^3 + 3x^2y + 3xy^2 + y^3) + y(x^3 + 3x^2y + 3xy^2 + y^3)\\ & = x^4 + 3x^3y + 3x^2y^2 + xy^3 + x^3y + 3x^2y^2 + 3xy^3 + y^4\\ & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \end{align*}
Now, let's look at what the theorem is saying: If $n = 3$, we have $$(x + y)^3 = (x + y)(x + y)(x + y)$$ Each term of the product is formed by selecting either an $x$ or $y$ from each of the three factors.
There are $2^3 = 8$ possibilities: $$ \begin{array}{c c c c} \text{first factor} & \text{second factor} & \text{third factor} & \text{product} & \text{simplified product}\\ x & x & x & xxx & x^3\\ x & x & y & xxy & x^2y\\ x & y & x & xyx & x^2y\\ x & y & y & xyy & xy^2\\ y & x & x & yxx & x^2y\\ y & x & y & yxy & xy^2\\ y & y & x & yyx & xy^2\\ y & y & y & yyy & y^3 \end{array} $$ Adding the terms yields $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.
If $n = 4$, then $$(x + y)^4 = (x + y)(x + y)(x + y)(x + y)$$ To form, the product, we take either an $x$ or a $y$ from each of the four factors, which gives us $2^4 = 16$ terms.
$$ \begin{array}{c c c c} \text{first factor} & \text{second factor} & \text{third factor} & \text{fourth factor} & \text{product} & \text{simplified product}\\ x & x & x & x & xxxx & x^4\\ x & x & x & y & xxxy & x^3y\\ x & x & y & x & xxyx & x^3y\\ x & x & y & y & xxyy & x^2y^2\\ x & y & x & x & xyxx & x^3y\\ x & y & x & y & xyxy & x^2y^2\\ x & y & y & x & xyyx & x^2y^2\\ x & y & y & y & xyyy & xy^3\\ y & x & x & x & yxxx & x^3y\\ y & x & x & y & yxxy & x^2y^2\\ y & x & y & x & yxyx & x^2y^2\\ y & x & y & y & yxyy & xy^3\\ y & y & x & x & yyxx & x^2y^2\\ y & y & x & y & yyxy & xy^3\\ y & y & y & x & yyyx & xy^3\\ y & y & y & y & yyyy & y^4 \end{array} $$ Adding the terms yields $(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$.
Furthermore, \begin{align*} (x + y)^3 & = x^3 + 3x^2y + 3xy^2 + y^3\\ & = \binom{3}{3}x^3y^0 + \binom{3}{2}x^2y + \binom{3}{1}xy^2 + \binom{3}{0}x^0y^3\\ & = \sum_{k = 0}^{3} \binom{3}{k}x^ky^{3 - k} \end{align*} and \begin{align*} (x + y)^4 & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\\ & = \binom{4}{4}x^4y^0 + \binom{4}{3}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{1}x^1y^3 + \binom{4}{0}x^0y^4\\ & = \sum_{k = 0}^{4} \binom{4}{k}x^ky^{4 - k} \end{align*} Notice that the coefficient of $x^ky^{n - k}$ is equal to the number of times the factor $x$ appears in the product $x^ky^{n - k}$, which is the number of ways we can select $k$ factors of $x$ and $n - k$ factors of $y$ from the $n$ factors in the product $(x + y)^n$.