A binomial is an algebraic expression that is the sum of two terms, for example $a+b$. The problem of the binomial expansion is to write a power $(a+b)^n$, where $n \in \mathbb{N}$, as a polynomial in $a$ and $b$. For small exponents this is done by direct multiplication, e.g. $(a+b)^2 = a^2 + 2ab + b^2$ and $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$.
For any $n \in \mathbb{N}$ the binomial theorem states $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}.$$ The expansion has exactly $n+1$ terms; as $k$ increases from $0$ to $n$, the exponent of $a$ decreases from $n$ to $0$ while the exponent of $b$ increases from $0$ to $n$, and in each term the sum of the exponents equals $n$.
The binomial coefficients are defined by $$\binom{n}{k} = \frac{n!}{k!\,(n-k)!}, \quad 0 \le k \le n.$$ They obey Pascal's rule $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, which lets one build Pascal's triangle: each entry is the sum of the two entries above it. The rows $1;\ 1\ 1;\ 1\ 2\ 1;\ 1\ 3\ 3\ 1$ give exactly the coefficients of $(a+b)^0,(a+b)^1,(a+b)^2,(a+b)^3$.
The term of index $k$ (counting from $k=0$) in $(a+b)^n$ is $$T_{k+1} = \binom{n}{k} a^{n-k} b^{k}.$$ Thus the $(k+1)$-th term is obtained by choosing $k$. This formula is the main tool for finding a specific term, a required coefficient, or the constant (independent) term without writing the whole expansion.
Setting $a=b=1$ gives the sum of all coefficients: $$\sum_{k=0}^{n} \binom{n}{k} = 2^n.$$ Setting $a=1,\ b=-1$ gives $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$, so the sums of coefficients at even and odd positions are each $2^{n-1}$. The coefficients are symmetric, $\binom{n}{k} = \binom{n}{n-k}$, so the expansion reads the same from both ends.
Find the term containing $x^4$ in $(x+2)^6$. Using $T_{k+1} = \binom{6}{k} x^{6-k} 2^{k}$, we need $6-k=4$, hence $k=2$. Then $T_{3} = \binom{6}{2} x^{4} 2^{2} = 15 \cdot 4 \cdot x^4 = 60 x^4$. So the required term is $60 x^4$.
Find the constant term of $\left(x + \dfrac{1}{x}\right)^{8}$. The general term is $T_{k+1} = \binom{8}{k} x^{8-k} x^{-k} = \binom{8}{k} x^{8-2k}$. The term is constant when $8-2k=0$, i.e. $k=4$. Hence the constant term is $\binom{8}{4} = 70$.
Find the sum of the coefficients of the polynomial $(2x-3)^5$. The sum of coefficients of a polynomial $P(x)$ equals $P(1)$. Substituting $x=1$ gives $(2 \cdot 1 - 3)^5 = (-1)^5 = -1$. Therefore the sum of all coefficients is $-1$.
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