Binomial theorem — expansion, general term: Exercises

Before we start, one reminder about the counting number $(r n)$ — this is the engine of every answer.

Level 1 — Recognition

L1.1 How many terms are in the expansion of $(3 x - 5)^{11}$ ?

Recall Solution

A binomial $(a + b)^{n}$ always expands into == $n + 1$ == terms — one for each value of $r$ from $0$ up to $n$ . Here $n = 11$ , so the count is $n + 1 = 11 + 1 = 12 .$ What we did: we did not expand anything — we just read off $n$ and added $1$ . The bracket contents ( $3 x$ , $- 5$ ) never affect the number of terms.

L1.2 In $(a + b)^{7}$ , write down the exponents of $a$ and $b$ in the term where $r = 3$ .

Recall Solution

The general term is $T_{r + 1} = (r 7) a^{7 - r} b^{r}$ . With $r = 3$ :

exponent of $b$ is $r = 3$ ,
exponent of $a$ is $n - r = 7 - 3 = 4$ .

Check they sum to $n$ : $4 + 3 = 7 = n$ . ✓ The term is $(3 7) a^{4} b^{3} = 35 a^{4} b^{3}$ .

Level 2 — Application

L2.1 Find the coefficient of $x^{3}$ in $(1 + 2 x)^{6}$ .

Recall Solution

General term: $T_{r + 1} = (r 6) (1)^{6 - r} (2 x)^{r} = (r 6) 2^{r} x^{r}$ . We want the power of $x$ equal to $3$ , so $r = 3$ : $coefficient = (3 6) 2^{3} = 20 \times 8 = 160 .$ Why include $2^{r}$ ? Because $b = 2 x$ carries the number $2$ ; the numeric part $2^{r}$ multiplies the counting number $(r 6)$ . The coefficient is never just $(r n)$ when the letter carries a number.

L2.2 Write out the full expansion of $(2 + x)^{4}$ .

Recall Solution

Use $T_{r + 1} = (r 4) 2^{4 - r} x^{r}$ and let $r$ run $0 \to 4$ :

$r$	$(r 4)$	$2^{4 - r}$	term
0	1	16	$16$
1	4	8	$32 x$
2	6	4	$24 x^{2}$
3	4	2	$8 x^{3}$
4	1	1	$x^{4}$

$(2 + x)^{4} = 16 + 32 x + 24 x^{2} + 8 x^{3} + x^{4} .$ What we did: here $a = 2$ (a number) and $b = x$ . So the numeric factor $2^{4 - r}$ shrinks as $r$ grows, since $a$ 's power decreases.

L2.3 Find the $4$ th term in the expansion of $(x + 3)^{8}$ .

Recall Solution

The $4$ th term means $T_{r + 1}$ with $r + 1 = 4$ , i.e. $r = 3$ (the counter is one behind the term number): $T_{4} = (3 8) x^{8 - 3} 3^{3} = 56 \cdot x^{5} \cdot 27 = 1512 x^{5} .$ Why $r = 3$ ? Because $T_{1}$ is the case $r = 0$ . So the $4$ th term always uses $r = 3$ .

Level 3 — Analysis

L3.1 Find the term containing $x^{10}$ in $(x^{3} + \frac{2}{x})^{6}$ .

Recall Solution

General term: $T_{r + 1} = (r 6) (x^{3})^{6 - r} (\frac{2}{x})^{r} = (r 6) 2^{r} x^{3 (6 - r)} x^{- r} = (r 6) 2^{r} x^{18 - 4 r} .$ Why combine the exponents? So the whole term is a single power of $x$ — the only quantity we can set equal to a target. Set the exponent to $10$ : $18 - 4 r = 10 \Rightarrow 4 r = 8 \Rightarrow r = 2.$ $T_{3} = (2 6) 2^{2} x^{10} = 15 \cdot 4 x^{10} = 60 x^{10} .$

L3.2 Find the term independent of $x$ (the constant term) in $(2 x^{2} - \frac{1}{x})^{9}$ .

Recall Solution

General term: $T_{r + 1} = (r 9) (2 x^{2})^{9 - r} (\frac{- 1}{x})^{r} = (r 9) 2^{9 - r} (- 1)^{r} x^{2 (9 - r) - r} .$ The exponent of $x$ is $2 (9 - r) - r = 18 - 3 r$ . "Independent of $x$ " means this exponent is $0$ : $18 - 3 r = 0 \Rightarrow r = 6.$ $T_{7} = (6 9) 2^{3} (- 1)^{6} = 84 \cdot 8 \cdot 1 = 672 .$ Why keep $(- 1)^{r}$ ? Because $b = - \frac{1}{x}$ is negative. Since $r = 6$ is even, $(- 1)^{6} = + 1$ , so the sign stays positive here — but had $r$ been odd it would flip. Never drop the sign.

Level 4 — Synthesis

L4.1 Find the middle term of $(3 x - \frac{1}{2 x})^{8}$ .

Recall Solution

Since $n = 8$ is even, there is a single middle term at position $\frac{n}{2} + 1 = 5$ , which uses $r = 4$ . $T_{5} = (4 8) (3 x)^{8 - 4} (\frac{- 1}{2 x})^{4} = (4 8) 3^{4} x^{4} \cdot \frac{( - 1 ) ^{4}}{2 ^{4} x ^{4}} .$ The $x^{4}$ in the numerator and denominator cancel, leaving a constant: $T_{5} = 70 \cdot 81 \cdot \frac{1}{16} = \frac{5670}{16} = \frac{2835}{8} .$ What we combined: the middle-term rule ( $n$ even ⇒ one middle term at $\frac{n}{2} + 1$ ) and the algebra of a term whose $x$ -powers cancel. $\frac{2835}{8} = 354.375$ .

L4.2 In $(1 + x)^{n}$ , the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $n$ .

Recall Solution

In $(1 + x)^{n}$ the coefficient of $x^{r}$ is exactly $(r n)$ (no numeric factors, since both letters are "clean": $1$ and $x$ ). Setting the two coefficients equal: $(2 n) = (3 n) .$ Write them with factorials and simplify (divide both sides by $\frac{n !}{2 ! 3 ! \dots}$ style — easiest is the ratio): $\frac{( 3 n )}{( 2 n )} = \frac{n - 2}{3} = 1 \Rightarrow n - 2 = 3 \Rightarrow n = 5 .$ Check: $(2 5) = 10$ and $(3 5) = 10$ . ✓ Equal, as required. What we synthesised: the identity "coefficient of $x^{r}$ in $(1 + x)^{n}$ is $(r n)$ " from L2 plus the factorial ratio $(r + 1 n) / (r n) = \frac{n - r}{r + 1}$ from Combinations nCr.

Level 5 — Mastery

L5.1 Find the coefficient of $x^{5}$ in the product $(1 + x)^{4} (1 - x)^{6}$ .

Recall Solution

Strategy: we don't need every term — only the ways two chosen powers of $x$ multiply to give $x^{5}$ . Write the coefficient of $x^{i}$ in $(1 + x)^{4}$ as $(i 4)$ , and of $x^{j}$ in $(1 - x)^{6}$ as $(j 6) (- 1)^{j}$ . We need $i + j = 5$ , with $0 \leq i \leq 4$ and $0 \leq j \leq 6$ .

$i$	$j = 5 - i$	$(i 4)$	$(j 6) (- 1)^{j}$	product
0	5	1	$6 \cdot (- 1) = - 6$	$- 6$
1	4	4	$15 \cdot (+ 1) = 15$	$60$
2	3	6	$20 \cdot (- 1) = - 20$	$- 120$
3	2	4	$15 \cdot (+ 1) = 15$	$60$
4	1	1	$6 \cdot (- 1) = - 6$	$- 6$

Add the products: $- 6 + 60 - 120 + 60 - 6 = - 12 .$ What we did: the coefficient of $x^{5}$ in a product is the sum over all splits $i + j = 5$ of (coeff of $x^{i}$ ) $\times$ (coeff of $x^{j}$ ). Every negative sign came from the $(- 1)^{j}$ inside $(1 - x)^{6}$ — keeping it was essential.

L5.2 The first three terms of $(1 + k x)^{n}$ (with $k$ a constant, $n$ a positive integer) are $1 + 12 x + 60 x^{2} + \dots .$ Find $k$ and $n$ .

Recall Solution

Expand $(1 + k x)^{n}$ using the general term $T_{r + 1} = (r n) (k x)^{r} = (r n) k^{r} x^{r}$ :

$r = 0$ : constant $= 1$ ✓ (automatically),
$r = 1$ : coefficient of $x$ is $(1 n) k = nk$ ,
$r = 2$ : coefficient of $x^{2}$ is $(2 n) k^{2} = \frac{n ( n - 1 )}{2} k^{2}$ .

Match to the given expansion: $nk = 12 and \frac{n ( n - 1 )}{2} k^{2} = 60.$ Eliminate $k$ . From the first equation $k = \frac{12}{n}$ . Substitute into the second: $\frac{n ( n - 1 )}{2} \cdot \frac{144}{n ^{2}} = 60 \Rightarrow \frac{72 ( n - 1 )}{n} = 60.$ $72 (n - 1) = 60 n \Rightarrow 72 n - 72 = 60 n \Rightarrow 12 n = 72 \Rightarrow n = 6.$ Then $k = \frac{12}{6} = 2$ . $n = 6, k = 2.$ Check the $x^{2}$ term: $\frac{6 \cdot 5}{2} \cdot 2^{2} = 15 \cdot 4 = 60$ . ✓

L5.3 Find the greatest coefficient in the expansion of $(1 + x)^{10}$ , and state which term it belongs to.

Recall Solution

In $(1 + x)^{10}$ the coefficients are $(0 10), (1 10), \dots, (10 10)$ — the $10$ th row of Pascal's Triangle. These rise to a single peak at the middle and fall symmetrically. For even $n = 10$ the peak is at $r = \frac{n}{2} = 5$ : $(5 10) = \frac{10 !}{5 ! 5 !} = 252.$ This is the coefficient of $x^{5}$ , i.e. the $T_{6}$ term (since $r = 5 \Rightarrow r + 1 = 6$ ). $Greatest coefficient = 252, in T_{6} = 252 x^{5} .$ Why the middle? Because $\frac{( r + 1 10 )}{( r 10 )} = \frac{10 - r}{r + 1}$ is $> 1$ for $r < 4.5$ (coefficients still rising) and $< 1$ for $r > 4.5$ (falling). The turnover sits exactly at $r = 5$ . See the figure below.

Figure — Binomial theorem — expansion, general term

Active recall

Recall Quick self-test (predict, then reveal)

Which $r$ gives the constant term of $(2 x^{2} - \frac{1}{x})^{9}$ ? ::: $18 - 3 r = 0 \Rightarrow r = 6$ . How many terms in $(3 x - 5)^{11}$ ? ::: $12$ . Coefficient of $x^{3}$ in $(1 + 2 x)^{6}$ ? ::: $(3 6) 2^{3} = 160$ . Position of the middle term of $(1 + x)^{10}$ ? ::: The $6$ th term, $r = 5$ , coefficient $252$ . In $(1 + x)^{n}$ , if coeffs of $x^{2}$ and $x^{3}$ are equal, then $n =$ ? ::: $5$ .

Connections

Binomial theorem — expansion, general term — the parent: where $T_{r + 1}$ comes from.
Combinations nCr — the ratio $(r + 1 n) / (r n) = \frac{n - r}{r + 1}$ used in L4/L5.
Pascal's Triangle — the row- $n$ coefficients and their single peak.
Factorials — how each $(r n)$ is computed.
Binomial Probability Distribution — same coefficients power a random experiment.

Exercises — Binomial theorem — expansion, general term

Level 1 — Recognition

Level 2 — Application

Level 3 — Analysis

Level 4 — Synthesis

Level 5 — Mastery

Active recall

Connections

Solution Map