What is the expansion $(x+a)^b$? I don't have strong math background. Whats is the expansion of the following equation: $(x+a)^b$.
 A: This seems a difficult problem for someone without a strong mathematical background.  One way to approach this is to arrange $b$ copies of $(x+a)$, as follows:
$$
\underbrace{(x+a)(x+a) \cdots (x+a)}_{\text{$b$ copies}}
$$
The product of all of these sums $(x+a)$ will be a bunch of terms of the form $x^ka^{b-k}$, where $k$ is some integer between $0$ and $b$, inclusive.  Each of the copies of $(x+a)$ can contribute either a factor of $x$, or a factor of $a$.  We then count up the number of ways to choose $k$ factors of $x$ and $b-k$ factors of $a$, and that will give us the coefficient of $x^ka^{b-k}$.  This works, ultimately, because multiplication distributes over addition: $u(v+w) = uv+uw$.
So, for instance, there is only one way to choose the $x$ from each of $b$ copies of $x+a$; those $b$ factors of $x$ multiply to $x^b$, so the coefficient of $x^b$ is just $1$.  On the other hand, since there are $b$ copies, there are $b$ ways to choose the $a$ from one of those $b$ copies (and consequently choose the $x$ from the remaining $b-1$ copies), so the coefficient of $x^{b-1}a$ is $b$.  Therefore, the expansion of $(x+a)^b$ must begin
$$
x^b+bx^{b-1}a+\cdots
$$
We can continue along the rest of the terms in the same way.  It turns out that there is a fairly simple formula for the number of ways to choose $k$ factors of $x$ and $b-k$ factors of $a$ from a total of $b$ copies of $(x+a)$; that formula is
$$
\binom{b}{k} = \frac{b!}{k!(b-k)!}
$$
which is in fact read "$b$ choose $k$", and where $b!$ is read "$b$ factorial" and represents the product $b \times (b-1) \times (b-2) \times \cdots \times 2 \times 1$.  Thus, the overall formula for $(x+a)^b$ is
$$
\binom{b}{b}x^ba^0 + \binom{b}{b-1}x^{b-1}a^1 + \binom{b}{b-2}x^{b-2}a^2 +
\cdots + \binom{b}{1}x^1a^{b-1} + \binom{b}{0}x^0a^b
$$
which we write in shorthand as simply
$$
(x+a)^b = \sum_{k=0}^b \binom{b}{k} x^ka^{b-k}
$$
A: $$(x + a)^b = \sum^{b}_{k=0} {{b}\choose {k}}x^{b-k}a^{k}$$
Where $b \choose k$ is defined as $${r \choose k}:=\frac{r!}{(r-k)!k!}.$$
Proof: We can prove it by induction.
For $r=0$ we get:
$$1=(x+a)^0=\sum^{0}_{k=0}{0 \choose k}x^{0-k}a^k=\frac{b!}{b!0!}x^0x^0=1.$$
Assume that:
$$P(r) : (x + a )^r = \sum^{r}_{k=0} {{r}\choose {k}}x^{r-k}a^{k}.$$
It then follows that:
$$(x + a )^{r+1} = \left(\sum^{r}_{k=0} {{r}\choose {k}}x^{r-k}a^{k}\right)(x + a) = \left(\sum^{r}_{k=0} {{r}\choose {k}}x^{r-k + 1}a^{k}\right) + \left(\sum^{r}_{k=0} {{r}\choose {k}}x^{r-k}a^{k+1}\right) = x^{r+1}+\left(\sum^{r}_{k=1} \left({{r}\choose {k}}+{{r}\choose {k-1}}\right)x^{r+1-k}a^{k}\right)+a^{r+1}= (x+a)^{r+1}$$ 
