Binomial theorem for positive integral index

Theorem 5.1 (Binomial theorem for positive integral index). If $n$ is any positive integer, then

Proof (by mathematical induction). Let $P(n)$ be the stated equality. Since $^1C_0={}^1C_1=1$, $P(1)$ reads $a^1b^0+a^0b^1=a+b$, which is exactly $(a+b)^1$ — so $P(1)$ is true. Assume $P(k)$ holds for some positive integer $k$. Multiplying both sides by $(a+b)$ and distributing over the two copies of the expansion, the coefficient of $a^{k-r+1}b^r$ that emerges is $^kC_r+{}^kC_{r-1}$ — and the identity $^kC_r+{}^kC_{r-1}={}^{k+1}C_r$ (proved in Chapter 4) turns this exactly into the $P(k+1)$ statement. So $P(k)\Rightarrow P(k+1)$, and by induction $P(n)$ holds for every $n\in\mathbb N$. $\blacksquare$

Standing remarks.

1. The expansion may equally be written $\displaystyle(a+b)^n=\sum_{k=0}^n {}^nC_k\,a^{n-k}b^k=\sum_{k=0}^n {}^nC_k\,a^kb^{n-k}$.
2. $(a+b)^n$, $n\in\mathbb N$, contains exactly $n+1$ terms.
3. As one reads left to right, the power of $a$ decreases by $1$ each term while the power of $b$ increases by $1$; the two powers always add up to $n$.
4. The $(r+1)^{th}$ term, called the general term, is $T_{r+1}={}^nC_r\,a^{n-r}b^r$, for $r=0,1,\ldots,n$.
5. Combinatorial meaning. In the product $(a+b)(a+b)\cdots(a+b)$ ($n$ factors), to pick up $b^r$ one must choose $b$ from exactly $r$ of the $n$ factors — and there are $^nC_r$ ways to make that choice, which is exactly why $^nC_r$ is the coefficient of $a^{n-r}b^r$.
6. Symmetric coefficients. Coefficients equidistant from the two ends are equal, since $^nC_r={}^nC_{n-r}$. …