Binomial Theorem for a Positive Integral Index and Pascal's Triangle

Expanding $(a+b)^2$ or $(a+b)^3$ by hand is easy, but $(a+b)^{10}$ would mean multiplying ten brackets together. The Binomial Theorem gives the entire expansion of $(a+b)^n$ for any positive integer $n$ in one stroke, without doing the multiplication.

The numbers $\binom{n}{r}={}^{n}C_r=\dfrac{n!}{r!\,(n-r)!}$ are the binomial coefficients. Read the expansion column by column: the power of $a$ falls from $n$ down to $0$, the power of $b$ rises from $0$ up to $n$, and in every single term the two powers add to $n$. There are exactly $n+1$ terms.

Reading the formula correctly. The two symbols $a$ and $b$ are the whole first and second entries of the bracket — including any sign, number or power they carry. In $(2x-3)^n$ you must take $a=2x$ and $b=-3$, never just $x$ and $3$; the minus sign rides along with $b$, so $b^r=(-3)^r$ alternates the signs of the expansion. Forgetting to carry the coefficient or the sign is the single most common slip in this whole chapter.

Properties worth memorising: The expansion is symmetric because $\binom{n}{r}=\binom{n}{n-r}$, so coefficients read the same forwards and backwards. Putting $a=b=1$ gives $\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$ (the sum of all coefficients), while $a=1,\,b=-1$ gives the alternating sum $\binom{n}{0}-\binom{n}{1}+\cdots=0$. Adding and subtracting those two results shows the sum of the even-position coefficients equals the sum of the odd-position ones, each being $2^{n-1}$. A handy special case is $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^{r}$.

Pascal's Triangle generates these coefficients without any factorials. Each row begins and ends in $1$, and every interior number is the sum of the two numbers diagonally above it — the identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$, known as Pascal's rule. To find row $n$ you only need row $n-1$, so the triangle is a quick hand method whenever $n$ is small (say $n\le 8$); for larger $n$ the factorial formula is faster.

Using the theorem to evaluate numbers. A power like $(1.01)^5$ or $98^3$ becomes easy once you write the base as a sum or difference of a round number and a small number, then apply the expansion. Write $1.01=1+0.01$ and $98=100-2$. Because each later term carries a higher power of the small piece, the terms shrink fast — so for an approximation you may keep only the first two or three terms, while for an exact value you keep them all. This is why $(1+x)^n\approx 1+nx$ for very small $x$: every term from $x^2$ onward is negligible.

Deeper Insight — why the coefficients are exactly $ {}^{n}C_r$: The binomial coefficient is not a mysterious lookup number; it is a count. When you expand $(a+b)^n=(a+b)(a+b)\cdots(a+b)$ with $n$ brackets, every term in the product is formed by choosing either $a$ or $b$ from each bracket. A term $a^{n-r}b^{r}$ arises whenever you pick $b$ from exactly $r$ of the $n$ brackets and $a$ from the rest — and the number of ways to choose those $r$ brackets is precisely $\binom{n}{r}$, the number of $r$-element subsets of an $n$-element set. This is why Pascal's identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ holds: a subset either contains a fixed element (choose the other $r-1$ from $n-1$) or it does not (choose all $r$ from $n-1$). The same subset-counting also explains why all the coefficients add to $2^n$: that is the total number of subsets of an $n$-set. Seeing algebra and counting as two views of one idea makes the whole chapter feel inevitable rather than memorised.