Worked examples — General term of binomial expansion — finding specific terms

This is a child of the parent topic. If any word below feels unfamiliar, the prerequisites are Binomial Theorem for positive integer index, Combinations nCr and Factorials.

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The scenario matrix

Before solving anything, let us list every distinct kind of problem this topic can throw. Each row is a "cell". Every worked example below is tagged with the cell it fills, so together they cover the whole grid.

Cell	What makes it different	Example
A — plain "which term"	asks for the $k$-th term directly, no exponent-matching	Ex 1
B — coefficient of a power, positive exponents	both entries have $x$ to positive powers	Ex 2
C — coefficient with a negative-power entry	one entry is $1/x^m$, exponents subtract	Ex 3
D — term independent of $x$ (net power $=0$)	set the collected exponent to zero	Ex 4
E — sign-sensitive term	entry carries a minus, answer sign depends on parity of $r$	Ex 5
F — "no such term" / degenerate	the exponent equation gives a non-integer or out-of-range $r$	Ex 6
G — two variables, match BOTH exponents	like $x^p{y}^q$, one $r$ must satisfy two demands	Ex 7
H — real-world word problem	binomial hidden inside a story (probability-style count)	Ex 8
I — exam twist: ratio / relationship between terms	uses $T_{r+2}/T_{r+1}$	Ex 9

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Ex 1 — Cell A: plain "which term"

Forecast: Guess before reading — which value of $r$ gives the 6th term, and will the sign be positive or negative? (Hint: count the b's.)

Steps.

1. Identify $a=3x,b=-2,n=8$. Why this step? The general term needs to know exactly what $a$, $b$, $n$ are, including the sign on $b$.
2. The 6th term means $r+1=6$, so $r=5$. Why? Term number is $r+1$ because $r$ starts at $0$.
3. Substitute: $T_6=\left(\genfrac{}{}{0ex}{}{8}{5}\right)(3x{)}^{8-5}(-2{)}^5=\left(\genfrac{}{}{0ex}{}{8}{5}\right)(3x{)}^3(-2{)}^5.$ Why? Direct plug-in of the general term — no exponent-matching needed for "which term" questions.
4. Evaluate each piece: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=56$, $(3x{)}^3=27x^3$, $(-2{)}^5=-32$. $T_6=56\cdot 27x^3\cdot (-32)=-48384x^3.$

Verify: Powers add: $(n-r)+r=3+5=8=n$ ✔. The sign is negative because we raised $-2$ to an odd power ✔, matching the forecast.

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Ex 2 — Cell B: coefficient of a power, positive exponents

Forecast: Both entries carry a positive power of $x$. Guess: after you collect exponents, will the equation for $r$ be $3(6-r)+r=10$ or something else?

Steps.

1. $a=x^3,b=2x,n=6$. General term: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{6}{r}\right)(x^3{)}^{6-r}(2x{)}^r=\left(\genfrac{}{}{0ex}{}{6}{r}\right)2^r{x}^{3(6-r)}{x}^r.$ Why this step? Write powers separately so we can combine them (the C of C-P-S).
2. Collect the power of $x$: $x^{3(6-r)+r}=x^{18-3r+r}=x^{18-2r}.$ Why? Adding exponents on the same base gives one net exponent — this is what we match against.
3. Set it equal to 10 (the P step): $18-2r=10\Rightarrow 2r=8\Rightarrow r=4.$ Why 10? We want the term that carries $x^{10}$.
4. Coefficient $=\left(\genfrac{}{}{0ex}{}{6}{4}\right)2^4=15\cdot 16=240.$

Verify: Check $r=4$ is in range $0\le r\le 6$ ✔. Net exponent at $r=4$: $18-2(4)=10$ ✔.

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Ex 3 — Cell C: coefficient with a negative-power entry

Forecast: Now one entry is $3/x=3x^{-1}$ — a negative power. Predict: will collecting exponents give a subtraction, and could $r$ come out large?

Steps.

1. $a=x^2,b=3x^{-1},n=11$: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{11}{r}\right)(x^2{)}^{11-r}(3x^{-1}{)}^r=\left(\genfrac{}{}{0ex}{}{11}{r}\right)3^r{x}^{2(11-r)}{x}^{-r}.$ Why? Rewrite $\frac{3}{x}$ as $3x^{-1}$ so the second factor is a clean power of $x$ — otherwise you can't add exponents.
2. Collect: $x^{2(11-r)-r}=x^{22-2r-r}=x^{22-3r}.$ Why? Combine into one exponent (C).
3. Set equal to 7 (P): $22-3r=7\Rightarrow 3r=15\Rightarrow r=5.$
4. Coefficient $=\left(\genfrac{}{}{0ex}{}{11}{5}\right)3^5=462\cdot 243=112266.$

Verify: $r=5\in [0,11]$ ✔. Exponent at $r=5$: $22-15=7$ ✔.

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Ex 4 — Cell D: term independent of $x$

Forecast: "Independent of $x$" means the net power of $x$ is zero. Guess the sign of the answer — the second entry has a minus, so it depends on the parity of $r$.

Steps.

1. $a=2x^2,b=-x^{-1},n=9$: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{9}{r}\right)(2x^2{)}^{9-r}(-x^{-1}{)}^r=\left(\genfrac{}{}{0ex}{}{9}{r}\right)2^{9-r}(-1{)}^r{x}^{2(9-r)}{x}^{-r}.$ Why? Split off the constant $2$, the sign $(-1{)}^r$, and the powers of $x$ separately — this keeps the sign safe. (This dodges the parent note's third slip: dropping the minus when raising a $-\frac{1}{x^m}$ entry to the power $r$.)
2. Collect: $x^{2(9-r)-r}=x^{18-3r}.$
3. Set to $0$ (P): $18-3r=0\Rightarrow r=6.$ Why 0? $x^0=1$ carries no $x$ — that's what "independent of $x$" means.
4. Evaluate: $\left(\genfrac{}{}{0ex}{}{9}{6}\right)=84$, $2^{9-6}=2^3=8$, $(-1{)}^6=+1$. $T_7=84\cdot 8\cdot 1=672.$

Verify: $r=6\in [0,9]$ ✔. Exponent $18-3(6)=0$ ✔. Since $r=6$ is even, the sign is positive ✔.

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Ex 5 — Cell E: sign-sensitive term (odd $r$)

Forecast: Same "independent of $x$" idea, but here $r$ will turn out odd — so the sign flips negative. Predict it before computing.

Steps.

1. $a=x,b=-2x^{-2},n=9$: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{9}{r}\right)x^{9-r}(-2x^{-2}{)}^r=(\genfrac{}{}{0ex}{}{9}{r})(-2{)}^r{x}^{9-r}{x}^{-2r}.$ Why? Keep $(-2{)}^r$ together so the sign and the factor $2^r$ ride along with the parity of $r$.
2. Collect: $x^{9-r-2r}=x^{9-3r}.$
3. Set to $0$: $9-3r=0\Rightarrow r=3.$
4. Evaluate: $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, $(-2{)}^3=-8$. $T_4=84\cdot (-8)=-672.$

Verify: $r=3$ odd $\Rightarrow$ negative ✔ (forecast confirmed). Exponent $9-3(3)=0$ ✔.

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Ex 6 — Cell F: degenerate / "no such term"

Forecast: Before any algebra — guess "yes" or "no". A term exists only if the exponent equation gives an integer $r$ inside $[0,8]$.

Steps.

1. General term: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{8}{r}\right)(x^3{)}^{8-r}(x^{-1}{)}^r=\left(\genfrac{}{}{0ex}{}{8}{r}\right)x^{3(8-r)-r}=\left(\genfrac{}{}{0ex}{}{8}{r}\right)x^{24-4r}.$ Why? Collect the exponent as always.
2. For "independent of $x$": $24-4r=0\Rightarrow r=6.$ Why check? $r$ must be a whole number between $0$ and $n$; if it isn't, the term doesn't exist.
3. $r=6$ is an integer in $[0,8]$ — so the term exists. Coefficient $=\left(\genfrac{}{}{0ex}{}{8}{6}\right)=28.$

Now the genuinely degenerate twist: does ${\left(x^3+\frac{1}{x}\right)}^8$ contain $x^5$?

4. $24-4r=5\Rightarrow 4r=19\Rightarrow r=\mathrm{4.75.}$ Not an integer $\Rightarrow$ no such term; the coefficient of $x^5$ is $0$.

Verify: Every attainable exponent is $24-4r$ for $r=0,\dots ,8$, i.e. $\{24,20,16,12,8,4,0,-4,-8\}$ — all $\equiv 0(\mathrm{mod}4)$. Since $5\not\equiv 0(\mathrm{mod}4)$, $x^5$ genuinely cannot appear ✔.

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Ex 7 — Cell G: two variables, match BOTH exponents

Forecast: With two variables you can't just match one exponent — but note $4+6=10=n$, which is exactly why it's possible. Guess: does $r$ count the $x$'s or the $y$'s?

Steps.

1. Here $a=x,b=y,n=10$: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)x^{10-r}{y}^r.$ Why? The general term already separates the two variables cleanly.
2. Match both exponents at once: we need $x^{10-r}=x^4$ and $y^r=y^6$, i.e. $10-r=4$ and $r=6$. Why both? One value of $r$ must satisfy the whole monomial; luckily $10-6=4$, so it's consistent.
3. Since the two demands agree ($r=6$), the coefficient is $\left(\genfrac{}{}{0ex}{}{10}{6}\right)=210.$

Verify: Exponents sum to $n$: $4+6=10$ ✔ — the necessary condition for the term to exist. $\left(\genfrac{}{}{0ex}{}{10}{6}\right)=\left(\genfrac{}{}{0ex}{}{10}{4}\right)=210$ ✔ (symmetry of Combinations nCr).

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Ex 8 — Cell H: real-world word problem

Forecast: Map the story onto $(a+b{)}^n$. Guess: does "2 Tails" mean $r=2$, and is $t$ our "$b$"?

Steps.

1. Set $a=0.6h,b=0.4t,n=5$. "Exactly 2 Tails" means we pick $b=0.4t$ from $r=2$ brackets. Why? Each bracket is one toss; choosing $t$ from $r$ of them means $r$ Tails (parent's box analogy).
2. The term is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)(0.6h{)}^3(0.4t{)}^2=\left(\genfrac{}{}{0ex}{}{5}{2}\right)(0.6{)}^3(0.4{)}^2{h}^3{t}^2.$ Why $r=2$? We chose Tails twice; the remaining $5-2=3$ tosses are Heads automatically.
3. Numerical coefficient (the probability of exactly 2 Tails): $\left(\genfrac{}{}{0ex}{}{5}{2}\right)(0.6{)}^3(0.4{)}^2=10\cdot 0.216\cdot 0.16=\mathrm{0.3456.}$

Verify: It's a probability, so $0\le 0.3456\le 1$ ✔. Sum of ALL terms $=(0.6+0.4{)}^5=1^5=1$ ✔ — total probability is 1, confirming the model.

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Ex 9 — Cell I: exam twist, ratio of consecutive terms

Forecast: Equal consecutive coefficients happen near the greatest term (see Greatest coefficient and greatest term). For $(1+x{)}^{10}$ the biggest coefficient is $\left(\genfrac{}{}{0ex}{}{10}{5}\right)$. Guess: will the balance land at $r=4$ or $r=5$?

Steps.

1. With $a=1,b=x,n=10$: $T_{r+1}=\left(\genfrac{}{}{0ex}{}{10}{r}\right)x^r$ and $T_{r+2}=\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)x^{r+1}$. Why? Bump the index by 1 to get the next term.
2. Form the full term ratio, keeping the $x$-power factor: $\frac{T_{r+2}}{T_{r+1}}=\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)x^{r+1}}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)x^r}=\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)}\cdot x^{(r+1)-r}=\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)}\cdot x.$ Why keep the $x$? $T_{r+2}$ has one more power of $x$ than $T_{r+1}$, so the ratio genuinely carries a factor $x$. We must show it, not hide it.
3. Remove the $x$ factor explicitly by evaluating at $x=1$, which turns "equal terms" into "equal coefficients": ${\frac{T_{r+2}}{T_{r+1}}|}_{x=1}=\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)}\cdot 1^1=\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)}.$ Why $x=1$? Substituting $x=1$ makes $x^1=1$, so the $x$-factor disappears cleanly and we are left comparing the pure coefficients — exactly what "consecutive coefficients equal" asks.
4. Simplify the coefficient ratio. Why does it collapse so neatly? $\frac{\left(\genfrac{}{}{0ex}{}{10}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{10}{r}\right)}=\frac{\frac{10!}{(r+1)!(9-r)!}}{\frac{10!}{r!(10-r)!}}=\frac{r!(10-r)!}{(r+1)!(9-r)!}=\frac{10-r}{r+1}.$ Why? The $10!$ cancels; $\frac{r!}{(r+1)!}=\frac{1}{r+1}$ and $\frac{(10-r)!}{(9-r)!}=10-r$ (one factor survives each). This is the factorial cancellation from Factorials and Binomial coefficients and Pascal's Triangle.
5. Set the coefficient ratio equal to $1$: $\frac{10-r}{r+1}=1\Rightarrow 10-r=r+1\Rightarrow 2r=9\Rightarrow r=\mathrm{4.5.}$ Why? Equal consecutive coefficients need the ratio to be exactly $1$.
6. $r=4.5$ is not an integer — so no two consecutive coefficients are exactly equal; instead the coefficients rise up to $r=5$ then fall. The single greatest coefficient is $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$.

Verify: Neighbours of the peak: $\left(\genfrac{}{}{0ex}{}{10}{4}\right)=210$ and $\left(\genfrac{}{}{0ex}{}{10}{6}\right)=210$ (equal by symmetry), while $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252>210$ ✔ — confirming a single peak at $r=5$, and no equal-consecutive pair.

Picture: the single peak

The bar chart below plots each coefficient $\left(\genfrac{}{}{0ex}{}{10}{r}\right)$ against $r$ for $(1+x{)}^{10}$. Read it left to right: the mint bars climb, the coral bar at $r=5$ is the tallest (the single greatest coefficient $252$), then the bars fall back down symmetrically. The two lavender bars at $r=4$ and $r=6$ are equal ($210$ each) — they sit either side of the peak but are not adjacent to each other, which is why no two neighbouring bars have the same height. The dashed butter line marks the balance point $r=4.5$: because it lands between two integers, there is no integer $r$ where consecutive coefficients match.

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Connections

• Parent: finding specific terms
• Binomial Theorem for positive integer index
• Binomial coefficients and Pascal's Triangle
• Middle term of a binomial expansion
• Greatest coefficient and greatest term
• Combinations nCr
• Factorials

Concept Map