General term of binomial expansion — finding specific terms: Exercises

Read the figure below before starting. It is the map every solution on this page walks: the general-term machine on the left, then the three C-P-S moves, ending in the red answer box. Whenever a solution says "(C)", "(P)", "(S)", glance back here to see which arrow you are on.

Figure — General term of binomial expansion — finding specific terms

Figure s01 — the C-P-S pipeline. Left box: the general term you always write first. Middle boxes: Collect the power, Put the exponent equal to the target. Right/red box: Solve for $r$ and plug back to read off the term or coefficient. The red box is always the thing the question actually asks for.

Level 1 — Recognition (read off the machine)

L1.1

In the expansion of $(x + y)^{10}$ , write the general term $T_{r + 1}$ and state how many terms there are.

Recall Solution

Match to $(a + b)^{n}$ : here $a = x$ , $b = y$ , $n = 10$ . $T_{r + 1} = (r 10) x^{10 - r} y^{r} .$ Since $r$ runs $0 \to 10$ , there are 11 terms (that is $n + 1$ ). See Middle term of a binomial expansion for what "middle" means when there are $11$ terms.

L1.2

Which value of $r$ gives the 7th term of $(a + b)^{n}$ ? Write that term for $(x + 3)^{8}$ .

Recall Solution

Term number is $r + 1$ , so $r + 1 = 7 \Rightarrow r = 6$ . For $(x + 3)^{8}$ : $a = x$ , $b = 3$ , $n = 8$ : $T_{7} = (6 8) x^{2} 3^{6} = 28 \cdot x^{2} \cdot 729 = 20412 x^{2} .$

Level 2 — Application (collect the power, solve for $r$ )

L2.1

Find the coefficient of $x^{5}$ in $(2 + x)^{9}$ .

Recall Solution

$a = 2$ , $b = x$ , $n = 9$ : $T_{r + 1} = (r 9) 2^{9 - r} x^{r} .$ (C) The power of $x$ is already collected as just $r$ . (P) Put it equal to the target: $r = 5$ . (S) Solve — here $r$ is already isolated, so: $coefficient = (5 9) 2^{4} = 126 \cdot 16 = 2016.$

L2.2

Find the coefficient of $x^{7}$ in $(2 x - 3)^{10}$ .

Recall Solution

$a = 2 x$ , $b = - 3$ , $n = 10$ : $T_{r + 1} = (r 10) (2 x)^{10 - r} (- 3)^{r} = (r 10) 2^{10 - r} (- 1)^{r} 3^{r} x^{10 - r} .$ (C) Collected power of $x$ is $10 - r$ . (P) Put it equal to $7$ : $10 - r = 7$ . (S) Solve: $r = 3$ . $coefficient = (3 10) 2^{7} (- 1)^{3} 3^{3} = 120 \cdot 128 \cdot (- 1) \cdot 27 = - 414720.$ Note the sign came from $(- 1)^{3} = - 1$ — parity of $r$ decides it.

L2.3

Find the term containing $x^{9}$ in $(x^{2} + \frac{2}{x})^{9}$ .

Recall Solution

$a = x^{2}$ , $b = 2 x^{- 1}$ , $n = 9$ : $T_{r + 1} = (r 9) (x^{2})^{9 - r} (2 x^{- 1})^{r} = (r 9) 2^{r} x^{18 - 2 r} x^{- r} = (r 9) 2^{r} x^{18 - 3 r} .$ (C) Collected exponent: $18 - 3 r$ . (P) Put it equal to $9$ : $18 - 3 r = 9$ . (S) Solve: $r = 3$ . $T_{4} = (3 9) 2^{3} x^{9} = 84 \cdot 8 \cdot x^{9} = 672 x^{9} .$

Level 3 — Analysis (constant terms, ratios, two conditions)

L3.1

Find the term independent of $x$ in $(3 x - \frac{1}{2 x ^{2}})^{9}$ .

Recall Solution

$a = 3 x$ , $b = - \frac{1}{2} x^{- 2}$ , $n = 9$ : $T_{r + 1} = (r 9) (3 x)^{9 - r} (- \frac{1}{2} x^{- 2})^{r} = (r 9) 3^{9 - r} (- 1)^{r} 2^{- r} x^{9 - r} x^{- 2 r} .$ (C) Collected power of $x$ : $9 - r - 2 r = 9 - 3 r$ . (P) "Independent of $x$ " means put the exponent equal to $0$ : $9 - 3 r = 0$ . (S) Solve: $r = 3$ . $T_{4} = (3 9) 3^{6} (- 1)^{3} 2^{- 3} = 84 \cdot 729 \cdot (- 1) \cdot \frac{1}{8} = - \frac{84 \cdot 729}{8} = - \frac{61236}{8} = - \frac{15309}{2} .$ So the constant term is exactly $- \frac{15309}{2}$ (the fully reduced fraction; as a decimal this is $- 7654.5$ , but keep the exact form).

L3.2

In $(x + \frac{1}{x})^{2 n}$ , find the term independent of $x$ (in terms of $n$ ).

Recall Solution

$a = x$ , $b = x^{- 1}$ , index $2 n$ : $T_{r + 1} = (r 2 n) x^{2 n - r} x^{- r} = (r 2 n) x^{2 n - 2 r} .$ (C) Collected exponent: $2 n - 2 r$ . (P) Independent of $x$ : $2 n - 2 r = 0$ . (S) Solve: $r = n$ . $T_{n + 1} = (n 2 n) .$ This is exactly the middle (single) term because the index $2 n$ is even — see Middle term of a binomial expansion.

L3.3

Two conditions: the coefficients of $T_{2}$ and $T_{3}$ in $(1 + x)^{n}$ are in ratio $1 : 4$ . Find $n$ .

Recall Solution

$a = 1$ , $b = x$ : $T_{r + 1} = (r n) x^{r}$ , so coefficient of $T_{r + 1}$ is $(r n)$ . $T_{2}$ has $r = 1$ : coefficient $(1 n) = n$ . $T_{3}$ has $r = 2$ : coefficient $(2 n) = \frac{n ( n - 1 )}{2}$ . $\frac{( 1 n )}{( 2 n )} = \frac{n}{\frac{n ( n - 1 )}{2}} = \frac{2}{n - 1} = \frac{1}{4} .$ $\Rightarrow n - 1 = 8 \Rightarrow n = 9.$

Level 4 — Synthesis (build the condition yourself)

L4.1

Find $r$ so that the coefficients of $x^{r}$ and $x^{r + 1}$ in $(1 + x)^{14}$ are equal.

Recall Solution

Coefficient of $x^{k}$ in $(1 + x)^{14}$ is $(k 14)$ . We need $(r 14) = (r + 1 14)$ . Two binomial coefficients $(k n)$ and $(m n)$ are equal when $k = m$ (impossible here since $r \neq = r + 1$ ) or when $k + m = n$ (symmetry of Binomial coefficients and Pascal's Triangle). $r + (r + 1) = 14 \Rightarrow 2 r + 1 = 14 \Rightarrow r = 6.5.$ Not an integer — so no such $r$ exists for $n = 14$ . This is a real answer: equal adjacent coefficients require $n$ odd (so that $2 r + 1 = n$ has an integer solution).

L4.2

In $(2 x^{2} - \frac{1}{x})^{12}$ , find the term whose $x$ -exponent equals the term number's $r$ . (i.e. find $r$ with collected exponent $= r$ , then give the term.)

Recall Solution

$a = 2 x^{2}$ , $b = - x^{- 1}$ , $n = 12$ : $T_{r + 1} = (r 12) (2 x^{2})^{12 - r} (- x^{- 1})^{r} = (r 12) 2^{12 - r} (- 1)^{r} x^{24 - 2 r} x^{- r} = (r 12) 2^{12 - r} (- 1)^{r} x^{24 - 3 r} .$ (C) Collected exponent: $24 - 3 r$ . (P) Impose "collected exponent $= r$ ": $24 - 3 r = r$ . (S) Solve: $24 = 4 r \Rightarrow r = 6$ . $T_{7} = (6 12) 2^{6} (- 1)^{6} x^{6} = 924 \cdot 64 \cdot 1 \cdot x^{6} = 59136 x^{6} .$

L4.3

Find the coefficient of $x^{- 2}$ in $(x - \frac{2}{x ^{3}})^{10}$ , or prove it does not exist.

Recall Solution

$a = x$ , $b = - 2 x^{- 3}$ , $n = 10$ : $T_{r + 1} = (r 10) x^{10 - r} (- 2)^{r} x^{- 3 r} = (r 10) (- 2)^{r} x^{10 - 4 r} .$ (C) Collected exponent: $10 - 4 r$ . (P) Put equal to $- 2$ : $10 - 4 r = - 2$ . (S) Solve: $4 r = 12 \Rightarrow r = 3$ (integer, in range — good). $coefficient = (3 10) (- 2)^{3} = 120 \cdot (- 8) = - 960.$

Level 5 — Mastery (multi-step, prove/derive)

L5.1

The coefficients of three consecutive terms of $(1 + x)^{n}$ are $220, 495, 792$ . Find $n$ and the position of the first of these terms.

Recall Solution

Let the three be $(r - 1 n) = 220$ , $(r n) = 495$ , $(r + 1 n) = 792$ . Use the ratio of consecutive coefficients, which cancels factorials cleanly: $\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r} = \frac{495}{220} = \frac{9}{4} .$ $\frac{( r + 1 n )}{( r n )} = \frac{n - r}{r + 1} = \frac{792}{495} = \frac{8}{5} .$ From the first: $4 (n - r + 1) = 9 r \Rightarrow 4 n + 4 = 13 r$ . From the second: $5 (n - r) = 8 (r + 1) \Rightarrow 5 n - 8 = 13 r$ . Set equal: $4 n + 4 = 5 n - 8 \Rightarrow n = 12$ . Then $13 r = 4 (12) + 4 = 52 \Rightarrow r = 4$ . Check: $(3 12) = 220$ , $(4 12) = 495$ , $(5 12) = 792$ . ✔ So $n = 12$ and the first coefficient $(3 12)$ belongs to $T_{4}$ (the 4th term).

L5.2

Show that in $(3 x + \frac{1}{x})^{15}$ there is exactly one term free of $x$ , and find it.

Recall Solution

Write roots as powers: $a = x^{1/3}$ , $b = x^{- 1/2}$ , $n = 15$ : $T_{r + 1} = (r 15) x^{(15 - r) /3} x^{- r /2} = (r 15) x^{\frac{15 - r}{3} - \frac{r}{2}} .$ (C) Collect the exponent over a common denominator $6$ : $\frac{15 - r}{3} - \frac{r}{2} = \frac{2 ( 15 - r ) - 3 r}{6} = \frac{30 - 5 r}{6} .$ (P) Free of $x$ : $30 - 5 r = 0$ . (S) Solve: $r = 6$ — a unique integer in $[0, 15]$ , so exactly one such term. $T_{7} = (6 15) = 5005.$

L5.3

Find the ratio of the coefficient of $x^{10}$ to the coefficient of $x^{7}$ in $(1 + x)^{15}$ , and interpret it.

Recall Solution

Coefficient of $x^{k}$ is $(k 15)$ : $\frac{( 10 15 )}{( 7 15 )} = \frac{3003}{6435} = \frac{7}{15} .$ Interpretation: $(7 15) = 6435$ is a peak (near the middle of row $15$ ), while $(10 15)$ sits farther out on the tail, so the ratio is less than $1$ . This is the symmetry-and-decay shape of Binomial coefficients and Pascal's Triangle and underlies Greatest coefficient and greatest term.

Connections

General term of binomial expansion — finding specific terms (index 3.3.13) (parent)
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

Coefficient of

x^{7}

(2 x - 3)^{10}

(3 10) 2^{7} (- 1)^{3} 3^{3} = - 414720

(from

r = 3

Term independent of

x

(x + 1/ x)^{2 n}

(n 2 n)

(from

r = n

Three consecutive coefficients

220, 495, 792

(1 + x)^{n}

n = 12

, starting at

T_{4}

Term free of

x

(3 x + 1/ x)^{15}

(6 15) = 5005

(from

r = 6

Coefficient of

x^{- 2}

(x - 2/ x^{3})^{10}

(3 10) (- 2)^{3} = - 960

(from

r = 3

Constant term in

(3 x - 1/ (2 x^{2}))^{9}

- \frac{15309}{2}

(from

r = 3