Coefficients & Specific Terms in the Expansion

For a positive integer $n$,

• It has exactly **$n+1$ terms**.
• The general term (the $(r+1)$-th term) is $T_{r+1} = \binom{n}{r} a^{\,n-r} b^{\,r}$, with $r = 0, 1, \ldots, n$.
• The exponents of $a$ and $b$ **always sum to $n$**; the powers of $a$ decrease while the powers of $b$ increase.

This single formula answers "find the term with $x^k$", "find the middle term", "find the constant term" — substitute and solve for $r$.

The binomial coefficient is

• Symmetry: $\binom{n}{r} = \binom{n}{n-r}$ (so $\binom{9}{4} = \binom{9}{5}$).
• Ends: $\binom{n}{0} = \binom{n}{n} = 1$.
• Pascal's rule: $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ (each entry is the sum of the two above it).
• Equal coefficients: $\binom{n}{a} = \binom{n}{b} \iff a = b \text{ or } a + b = n$ — the symmetry property in disguise, and a recurring NDA shortcut.

For an expansion in $x$, write $T_{r+1}$, simplify the exponent of $x$ to a single linear expression in $r$, set it equal to the target power, and solve:

• **Coefficient of $x^k$:** solve (exponent of $x$) $= k$ for $r$, then evaluate $T_{r+1}$.
• A term from the end: the $m$-th term from the end of an $(n+1)$-term expansion is the $(n+2-m)$-th from the start.
• If solving gives a non-integer $r$, that power simply does not appear (its coefficient is 0).

For $(a+b)^n$ (which has $n+1$ terms):

• **$n$ even:** a single middle term, the $\left(\tfrac{n}{2}+1\right)$-th term, $T_{\frac{n}{2}+1}$.
• **$n$ odd:** two middle terms, the $\left(\tfrac{n+1}{2}\right)$-th and $\left(\tfrac{n+3}{2}\right)$-th.

Tip: a square trinomial like $1 + 4x + 4x^2 = (1+2x)^2$ should be collapsed to a binomial first — then it has a clean middle term.

Write $T_{r+1}$, collect the exponent of $x$ into one expression in $r$, and set it to 0. Solve for $r$; that term is the constant. If $r$ comes out non-integer, there is no term independent of $x$.

Common condition types:

• Equal coefficients of two terms: equate the two general-term coefficients (use $\binom{n}{a}=\binom{n}{b} \iff a+b=n$).
• First three terms given (e.g. $1,\ 12x,\ 64x^2$): read off $\binom{n}{1}a = 12$ and $\binom{n}{2}a^2 = 64$, divide to eliminate, solve for $n$.
• Greatest coefficient of $(1+x)^n$: it is the middle coefficient $\binom{n}{\lfloor n/2 \rfloor}$.
• Sum of coefficients = value (e.g. $2^n = 256$): solve for $n$ first.

Simplify before counting:

• Conjugate product: $(a+b)(a-b) = a^2 - b^2$, so $(a+b)^k(a-b)^k = (a^2-b^2)^k$ has $k+1$ terms.
• Perfect-square trinomial: $1 + 2x + x^2 = (1+x)^2$; collapse, then count.
• Genuine trinomial $(a+b+c)^n$: the number of distinct terms is $\binom{n+2}{2}$.
• Sum/difference of two expansions $(a+b)^n \pm (a-b)^n$: like powers either add or cancel — count only the survivors.

• Rational terms: in $(p^{1/j} + q^{1/k})^n$, the general term carries exponents $\tfrac{(n-r)}{j}$ and $\tfrac{r}{k}$. A term is rational iff BOTH are integers; count the $r$ in $\{0,\ldots,n\}$ satisfying both divisibility conditions.
• General index: for any real $m$ and $|x|<1$, $(1+x)^m = \sum_{r\ge 0} \binom{m}{r} x^r$ with $\binom{m}{r} = \dfrac{m(m-1)\cdots(m-r+1)}{r!}$. This is how $(1 - \tfrac{x^2}{20})^{-1}$ becomes a geometric-type series.