Key Points

The expansion of $(a+b)^n$ for any positive integer n is $(a+b)^n = {}^n\mathrm{C}_0 a^n + {}^n\mathrm{C}_1 a^{n-1}b + {}^n\mathrm{C}_2 a^{n-2}b^2 + \dots + {}^n\mathrm{C}_n b^n$.

The binomial theorem can be expressed concisely using sigma notation as $(a+b)^n = \sum_{k=0}^{n} {}^n\mathrm{C}_k a^{n-k}b^k$.

The coefficients ${}^n\mathrm{C}_r$ are known as binomial coefficients. They are calculated using the formula ${}^n\mathrm{C}_r = \frac{n!}{r!(n-r)!}$ where $0 \le r \le n$.

Pascal's triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The numbers in the row corresponding to index $n$ are the binomial coefficients ${}^n\mathrm{C}_0, {}^n\mathrm{C}_1, \dots, {}^n\mathrm{C}_n$.

The total number of terms in the expansion of a binomial $(a+b)^n$ is always $n+1$, which is one more than the index $n$.

In successive terms of the expansion of $(a+b)^n$, the power of 'a' decreases by 1 (from $n$ down to 0), while the power of 'b' increases by 1 (from 0 up to $n$).

In each term of the expansion of $(a+b)^n$, the sum of the powers of 'a' and 'b' is constant and equal to the index $n$.

The expansion for a difference is $(x-y)^n = {}^n\mathrm{C}_0 x^n - {}^n\mathrm{C}_1 x^{n-1}y + {}^n\mathrm{C}_2 x^{n-2}y^2 - \dots + (-1)^n {}^n\mathrm{C}_n y^n$. The terms have alternating signs.

A common special case is the expansion of $(1+x)^n$, which is given by $(1+x)^n = {}^n\mathrm{C}_0 + {}^n\mathrm{C}_1 x + {}^n\mathrm{C}_2 x^2 + \dots + {}^n\mathrm{C}_n x^n$.

The sum of the binomial coefficients for any power $n$ is $2^n$. This is derived by setting $x=1$ in the expansion of $(1+x)^n$, resulting in ${}^n\mathrm{C}_0 + {}^n\mathrm{C}_1 + {}^n\mathrm{C}_2 + \dots + {}^n\mathrm{C}_n = 2^n$.

The sum of binomial coefficients with alternating signs is zero. This is shown by setting $x=1$ in the expansion of $(1-x)^n$, giving ${}^n\mathrm{C}_0 - {}^n\mathrm{C}_1 + {}^n\mathrm{C}_2 - \dots + (-1)^n {}^n\mathrm{C}_n = 0$.

The theorem is used to evaluate high powers of numbers by expressing the base as a sum or difference. For example, $(98)^5$ can be computed by expanding $(100-2)^5$.

Binomial expansion can be used to prove divisibility properties. For instance, expanding $6^n = (1+5)^n$ helps to show that $6^n - 5n$ leaves a remainder of 1 when divided by 25.

The theorem helps in comparing large numbers by expanding one of them. For example, expanding $(1.01)^{1000000} = (1+0.01)^{1000000}$ and examining the first few terms proves it is larger than 10,000.