Binomial Theorem for Positive Integral Index

Multiplying out $(a+b)^n$ directly for a large $n$ by repeated multiplication is slow and error-prone. The Binomial Theorem gives a direct formula for every term of the expansion, and the coefficients that appear are — not by coincidence — exactly the combination values $^{n}C_{r}$ from the previous section, which is why this topic sits naturally at the end of the same chapter.

Statement of the theorem

For any positive integer $n$,

Why the coefficients are $^{n}C_r$. Expanding $(a+b)^n = (a+b)(a+b)\cdots(a+b)$ ($n$ factors) means choosing either $a$ or $b$ from each of the $n$ factors and multiplying the choices together. The term containing $b^r$ (and therefore $a^{n-r}$) arises from every way of choosing $r$ of the $n$ factors to contribute a $b$ — and the number of ways to choose $r$ factors out of $n$, with order not mattering, is exactly $^{n}C_{r}$.

The general term

The $(r+1)$-th term of the expansion (note: the $(r+1)$-th term, not the $r$-th, because the count of $b$'s starts at $r=0$) is

This is the formula used whenever a question asks for "the $k$-th term" or "the term containing a specific power" without requiring the full expansion.

Number of terms and the middle term

• The expansion of $(a+b)^n$ has exactly $n+1$ terms (for $r = 0, 1, 2, \dots, n$).
• If $n$ is even, there is a single middle term: the $\left(\dfrac{n}{2}+1\right)$-th term.
• If $n$ is odd, there are two middle terms: the $\left(\dfrac{n+1}{2}\right)$-th and $\left(\dfrac{n+3}{2}\right)$-th terms.

Useful special results (setting $a=1,\,b=1$ or $b=-1$)

• Sum of all binomial coefficients: $^{n}C_0 + {}^{n}C_1 + \cdots + {}^{n}C_n = 2^{n}$ (put $a=b=1$).
• Alternating sum: $^{n}C_0 - {}^{n}C_1 + {}^{n}C_2 - \cdots = 0$ for $n \ge 1$ (put $a=1,\,b=-1$).

Pascal's Triangle …