Many exam questions never ask for a full expansion at all. They ask for one targeted piece — the term free of $x$, the coefficient of a specific power, or the numerically greatest term. Each is solved by squeezing the right value of $r$ out of the general term.
$$T_{r+1}=\binom{n}{r}a^{n-r}b^{r}\quad\Longrightarrow\quad\text{find the }r\text{ that gives the power you want.}$$
Term independent of $x$ (the constant term): Write the general term, collect every power of $x$ into a single exponent, set that exponent equal to $0$, and solve for $r$. If that $r$ is a non-negative integer $\le n$, the constant term is $T_{r+1}$ for that $r$; if it is not an integer, no constant term exists.
Coefficient of a given power $x^m$: Same idea — set the combined exponent of $x$ equal to $m$, solve for $r$, then read off $\binom{n}{r}$ times the numerical part. Distinguish the coefficient (the number multiplying $x^m$) from the binomial coefficient $\binom{n}{r}$, since powers of constants inside the bracket also contribute.
Ratios of consecutive coefficients. Neighbouring binomial coefficients are linked by a clean ratio that avoids recomputing factorials:
$$\dfrac{\binom{n}{r}}{\binom{n}{r-1}}=\dfrac{n-r+1}{r}.$$
This single relation answers questions like "in what ratio are three consecutive coefficients?" or "if consecutive coefficients are in the ratio $1:2:3$, find $n$" — set up the ratio twice and solve the simultaneous equations.
Terms equidistant from the ends. Because $\binom{n}{r}=\binom{n}{n-r}$, the binomial coefficient of the $(r+1)$th term from the beginning equals that of the $(r+1)$th term from the end. The terms themselves are equal only when $a=b$, but their coefficients always match — a fact that shortens many proofs.
Numerically greatest term in $(1+x)^n$ (for a fixed numerical $x$): consecutive terms are compared through their ratio
$$\dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\,|x|.$$
The terms increase while this ratio exceeds $1$ and decrease once it drops below $1$, so the greatest term sits where the ratio crosses $1$. Find the largest $r$ with $T_{r+1}\ge T_r$ — i.e. the integer part of $\dfrac{(n+1)|x|}{1+|x|}$ — and that locates the numerically greatest term.
Combined and trinomial-type expansions. A product such as $(1+x)^m(1+x)^n=(1+x)^{m+n}$ lets you read a coefficient straight off the merged power, and comparing it with the product of two separate expansions proves identities like $\sum_{k}\binom{m}{k}\binom{n}{p-k}=\binom{m+n}{p}$ (Vandermonde). A trinomial $(a+b+c)^n$ can be handled by grouping two of the symbols, e.g. $\big((a+b)+c\big)^n$, then expanding twice. These "combined binomial proofs" are nothing more than the same general term applied in two layers.
Deeper Insight — every special term is the same one question: It looks like several separate techniques, but constant terms, coefficient-hunting, ratios and greatest terms are all one move: control the exponent of $x$, then evaluate the binomial coefficient that subset-counting provides. Because $T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$ packs every power into a function of the single counter $r$, you are really solving a small linear equation (or inequality) in $r$ to pick the term you care about. The constant term is just "power $=0$", the coefficient question is "power $=m$", the ratio question fixes neighbouring $r$'s, and the greatest term swaps the equation for an inequality. Recognising this shared skeleton means you never memorise separate recipes — you set up the general term once and ask what value of $r$ the question is secretly demanding. The recurring $\binom{n}{r}$ is, as always, the count of how many of the $n$ brackets supplied the second symbol.
Find the term independent of $x$ in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{9}$.
Solution- General term: $T_{r+1}=\binom{9}{r}(x^2)^{9-r}\left(\dfrac{1}{x}\right)^{r}=\binom{9}{r}x^{18-2r}\cdot x^{-r}=\binom{9}{r}x^{18-3r}$.
- For the constant term set the exponent $18-3r=0\Rightarrow r=6$.
- Constant term $=\binom{9}{6}=\binom{9}{3}=84$.
Answer: The term independent of $x$ is $84$.
Find the coefficient of $x^6$ in the expansion of $\left(x-\dfrac{2}{x}\right)^{10}$ — careful, this needs $x^6$ specifically.
Solution- General term: $T_{r+1}=\binom{10}{r}x^{10-r}\left(-\dfrac{2}{x}\right)^{r}=\binom{10}{r}(-2)^{r}x^{10-2r}$.
- Set $10-2r=6\Rightarrow r=2$.
- Coefficient $=\binom{10}{2}(-2)^2=45\cdot 4=180$.
Answer: The coefficient of $x^6$ is $180$.
Find the term independent of $x$ in $\left(\dfrac{3}{2}x^2-\dfrac{1}{3x}\right)^{6}$.
Solution- $T_{r+1}=\binom{6}{r}\left(\dfrac{3}{2}x^2\right)^{6-r}\left(-\dfrac{1}{3x}\right)^{r}$; the power of $x$ is $2(6-r)-r=12-3r$.
- Set $12-3r=0\Rightarrow r=4$.
- $T_5=\binom{6}{4}\left(\dfrac{3}{2}\right)^{2}\left(-\dfrac{1}{3}\right)^{4}=15\cdot\dfrac{9}{4}\cdot\dfrac{1}{81}=15\cdot\dfrac{9}{324}=15\cdot\dfrac{1}{36}=\dfrac{5}{12}$.
Answer: The term independent of $x$ is $\dfrac{5}{12}$.
Find the coefficient of $x^{10}$ in the expansion of $(1+x^2)^{12}$.
Solution- General term: $T_{r+1}=\binom{12}{r}(x^2)^{r}=\binom{12}{r}x^{2r}$.
- Set $2r=10\Rightarrow r=5$.
- Coefficient $=\binom{12}{5}=792$.
Answer: The coefficient of $x^{10}$ is $792$.
Determine whether $\left(x+\dfrac{1}{x^2}\right)^{7}$ has a term independent of $x$.
Solution- $T_{r+1}=\binom{7}{r}x^{7-r}\left(\dfrac{1}{x^2}\right)^{r}=\binom{7}{r}x^{7-3r}$.
- For a constant term need $7-3r=0\Rightarrow r=\dfrac{7}{3}$.
- $\dfrac{7}{3}$ is not a whole number, so no value of $r$ works.
Answer: There is no term independent of $x$.
Find the numerically greatest term in the expansion of $(1+x)^{10}$ when $x=\dfrac{2}{3}$.
Solution- Locate the peak using $\dfrac{(n+1)|x|}{1+|x|}=\dfrac{11\cdot\frac{2}{3}}{1+\frac{2}{3}}=\dfrac{\frac{22}{3}}{\frac{5}{3}}=\dfrac{22}{5}=4.4$; the integer part is $4$, so the greatest is $T_{4+1}=T_5$ (here $r=4$).
- $T_5=\binom{10}{4}x^{4}=210\left(\dfrac{2}{3}\right)^{4}=210\cdot\dfrac{16}{81}=\dfrac{3360}{81}$.
- $\dfrac{3360}{81}=\dfrac{1120}{27}\approx 41.48$.
Answer: The numerically greatest term is $T_5=\dfrac{1120}{27}$.
The coefficients of three consecutive terms in $(1+x)^n$ are in the ratio $1:7:42$. Find $n$.
Solution- Let the three be $\binom{n}{r-1},\binom{n}{r},\binom{n}{r+1}$. Use $\dfrac{\binom{n}{r}}{\binom{n}{r-1}}=\dfrac{n-r+1}{r}=\dfrac{7}{1}$.
- And $\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac{n-r}{r+1}=\dfrac{42}{7}=6$.
- From the first, $n-r+1=7r\Rightarrow n+1=8r$. From the second, $n-r=6(r+1)\Rightarrow n=7r+6$.
- Substitute: $7r+6+1=8r\Rightarrow r=7$, hence $n=7(7)+6=55$.
Answer: $n=55$ (with $r=7$).
Show that the coefficients of the terms equidistant from the beginning and the end of $(1+x)^n$ are equal, and verify for the $3$rd terms from each end of $(1+x)^8$.
Solution- The $(r+1)$th term from the start has coefficient $\binom{n}{r}$; the $(r+1)$th from the end is the $(n-r+1)$th from the start, with coefficient $\binom{n}{n-r}$.
- Since $\binom{n}{r}=\binom{n}{n-r}$, the two coefficients are equal.
- Check $(1+x)^8$: 3rd term from start is $\binom{8}{2}=28$; 3rd from end is $\binom{8}{6}=28$.
Answer: Equidistant terms share the coefficient $\binom{n}{r}=\binom{n}{n-r}$; for $(1+x)^8$ both 3rd terms have coefficient $28$.
Find the coefficient of $x^5$ in the product $(1+x)^4(1+x)^6$.
Solution- Combine the powers: $(1+x)^4(1+x)^6=(1+x)^{10}$.
- The coefficient of $x^5$ in $(1+x)^{10}$ is $\binom{10}{5}$.
- $\binom{10}{5}=252$.
Answer: The coefficient of $x^5$ is $252$.
Find the numerically greatest term in $(2+3x)^9$ when $x=\dfrac{3}{2}$.
Solution- Write $(2+3x)^9=2^9\left(1+\dfrac{3x}{2}\right)^9$; with $x=\dfrac32$ the inner ratio is $\left|\dfrac{3x}{2}\right|=\dfrac{9}{4}$.
- Peak position: $\dfrac{(n+1)|t|}{1+|t|}$ with $t=\dfrac94$ gives $\dfrac{10\cdot\frac94}{1+\frac94}=\dfrac{\frac{90}{4}}{\frac{13}{4}}=\dfrac{90}{13}\approx 6.9$; integer part $6$, so the greatest term is $T_7$ ($r=6$).
- $T_7=2^9\binom{9}{6}\left(\dfrac94\right)^{6}=512\cdot 84\cdot\dfrac{531441}{4096}$. Note $\dfrac{512}{4096}=\dfrac18$, so $T_7=84\cdot\dfrac{531441}{8}=\dfrac{44641044}{8}=5580130.5$.
Answer: The numerically greatest term is $T_7=84\cdot\dfrac{9^6}{8}=\dfrac{44641044}{8}=5580130.5$.
Find the coefficient of $x^3$ in the expansion of $(1+x+x^2)^3$ (a trinomial) by grouping.
Solution- Group as $\big((1+x)+x^2\big)^3$ and expand by the binomial theorem in $x^2$: $\sum_{r=0}^{3}\binom{3}{r}(1+x)^{3-r}(x^2)^{r}$.
- An $x^3$ can come from $r=0$ (the $x^3$ term of $(1+x)^3$, coefficient $\binom{3}{3}=1$) or $r=1$ (factor $x^2$ times the $x^1$ term of $(1+x)^2$, coefficient $\binom31\binom21=3\cdot2=6$).
- Higher $r$ gives powers above $x^3$. Total coefficient $=1+6=7$.
Answer: The coefficient of $x^3$ is $7$.
Find the term independent of $x$ in $\left(2x^2-\dfrac{1}{x}\right)^{12}$.
Solution- $T_{r+1}=\binom{12}{r}(2x^2)^{12-r}\left(-\dfrac{1}{x}\right)^{r}=\binom{12}{r}2^{12-r}(-1)^{r}x^{24-2r-r}=\binom{12}{r}2^{12-r}(-1)^{r}x^{24-3r}$.
- Set $24-3r=0\Rightarrow r=8$.
- Term $=\binom{12}{8}2^{12-8}(-1)^{8}=495\cdot 16\cdot 1=7920$.
Answer: The term independent of $x$ is $7920$.