General and Middle Terms, with Mixed Practice, Worked Solutions

Ques­tion 1: Gen­eral term and the 4th term#

The prob­lem#

Write the gen­eral term of $(3x - y^2)^6$ and find the $4$th term.

Under­stand­ing the prob­lem#

The expres­sion is a bino­mial $(a + b)^n$ with first term $a = 3x$, sec­ond term $b = -y^2$ and power $n = 6$. The gen­eral term is a for­mula for any term of the expan­sion. The 4th term is one par­tic­u­lar term. Watch the sign: the sec­ond term is $-y^2$, not $y^2$.

The idea#

Use $T_{r+1} = {}^nC_r\, a^{n-r} b^r$. The 4th term is $T_{r+1}$ with $r + 1 = 4$, that is $r = 3$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $a = 3x$, $b = -y^2$, $n = 6$ into the gen­eral term.

$$T_{r+1} = {}^6C_r\,(3x)^{6-r}\,(-y^2)^r$$

This is the gen­eral term. You can also write it as $(-1)^r\,{}^6C_r\,3^{6-r}\,x^{6-r}\,y^{2r}$.

Step 2. For the 4th term put $r = 3$.

$$T_4 = {}^6C_3\,(3x)^{3}\,(-y^2)^3$$

Step 3. Work out each fac­tor sep­a­rately.

$$\displaystyle \begin{aligned} {}^6C_3 &= \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\\ (3x)^3 &= 27x^3\\ (-y^2)^3 &= -y^6 \end{aligned}$$

Step 4. Mul­ti­ply the fac­tors.

$$T_4 = 20 \times 27x^3 \times (-y^6) = -540\,x^3y^6$$

Check­ing the answer#

The pow­ers of $x$ and $y^2$ must add up to $6$: here $x^3$ and $(y^2)^3$, and $3 + 3 = 6$. The sign is neg­a­tive because an odd power ($3$) of $-y^2$ is taken.

Answer#

$T_{r+1} = {}^6C_r\,(3x)^{6-r}(-y^2)^r$, and $T_4 = -540\,x^3y^6$.

Com­mon mis­take to avoid#

Using $r = 4$ for the 4th term. The term num­ber is always one more than $r$.

Ques­tion 2: Coef­fi­cient of $x^4$#

The prob­lem#

Find the coef­fi­cient of $x^4$ in $(1 - 2x)^9$.

Under­stand­ing the prob­lem#

Here $a = 1$, $b = -2x$, $n = 9$. The coef­fi­cient of $x^4$ is the num­ber mul­ti­ply­ing $x^4$ in the expan­sion, includ­ing its sign.

The idea#

Write the gen­eral term, col­lect the power of $x$ in it, and choose $r$ so that this power is $4$.

Step-by-step solu­tion#

Step 1. Write the gen­eral term.

$$T_{r+1} = {}^9C_r\,(1)^{9-r}\,(-2x)^r = {}^9C_r\,(-2)^r\,x^r$$

Step 2. The power of $x$ is $r$. We need $x^4$, so $r = 4$.

Step 3. Find the coef­fi­cient with $r = 4$.

$$\displaystyle \begin{aligned} {}^9C_4 &= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\\ (-2)^4 &= 16\\ {}^9C_4\,(-2)^4 &= 126 \times 16 = 2016 \end{aligned}$$

Check­ing the answer#

An even power of $-2$ is pos­i­tive, so a pos­i­tive coef­fi­cient is right.

Answer#

The coef­fi­cient of $x^4$ is $2016$.

Ques­tion 3: Term free of $x$#

The prob­lem#

Find the term free of $x$ in $\displaystyle \left(2x + \frac{1}{3x^2}\right)^9$.

Under­stand­ing the prob­lem#

"Free of $x$" (also called "inde­pen­dent of $x$") means the term in which the power of $x$ is $0$, so it is just a num­ber. Here $a = 2x$, $\displaystyle b = \frac{1}{3x^2}$, $n = 9$.

The idea#

Write the gen­eral term, sim­plify the power of $x$ to a sin­gle expo­nent in $r$, and set that expo­nent equal to $0$.

Step-by-step solu­tion#

Step 1. Write the gen­eral term.

$$\displaystyle T_{r+1} = {}^9C_r\,(2x)^{9-r}\left(\frac{1}{3x^2}\right)^r$$

Step 2. Sep­a­rate the num­bers from the pow­ers of $x$.

$$\displaystyle \begin{aligned} T_{r+1} &= {}^9C_r\,2^{9-r}\,x^{9-r}\cdot\frac{1}{3^r}\,x^{-2r}\\ &= {}^9C_r\,\frac{2^{9-r}}{3^r}\,x^{9-3r} \end{aligned}$$

Step 3. Set the power of $x$ equal to zero.

$$9 - 3r = 0 \quad\Rightarrow\quad r = 3$$

So the required term is $T_4$.

Step 4. Put $r = 3$.

$$\displaystyle \begin{aligned} T_4 &= {}^9C_3\,\frac{2^{6}}{3^3}\\ &= 84 \times \frac{64}{27}\\ &= \frac{5376}{27} = \frac{1792}{9} \end{aligned}$$

(Here $\displaystyle {}^9C_3 = \frac{9 \times 8 \times 7}{6} = 84$, and $5376 \div 3 = 1792$, $27 \div 3 = 9$.)

Check­ing the answer#

With $r = 3$ the pow­ers are $x^{6}$ from $(2x)^6$ and $x^{-6}$ from $(x^{-2})^3$; they can­cel, as they should.

Answer#

The term free of $x$ is the 4th term, $\displaystyle \frac{1792}{9}$.

Ques­tion 4: Mid­dle terms#

The prob­lem#

Find the mid­dle term(s) of $\displaystyle \left(\frac{x}{3} + 3y\right)^6$ and of $(1 - x)^{11}$.

Under­stand­ing the prob­lem#

A bino­mial raised to the power $n$ has $n + 1$ terms. If $n$ is even there is one mid­dle term; if $n$ is odd there are two.

The idea#

For even $n$ the mid­dle term is $\displaystyle T_{\frac{n}{2}+1}$. For odd $n$ they are $\displaystyle T_{\frac{n+1}{2}}$ and $\displaystyle T_{\frac{n+3}{2}}$.

Step-by-step solu­tion#

Part (a): $\displaystyle \left(\frac{x}{3} + 3y\right)^6$

Step 1. Here $n = 6$ is even, so there are $7$ terms and one mid­dle term, $\displaystyle T_{\frac{6}{2}+1} = T_4$, which has $r = 3$.

Step 2. Write $T_4$.

$$\displaystyle T_4 = {}^6C_3\left(\frac{x}{3}\right)^{3}(3y)^3$$

Step 3. Sim­plify.

$$\displaystyle \begin{aligned} T_4 &= 20 \times \frac{x^3}{27} \times 27y^3\\ &= 20\,x^3y^3 \end{aligned}$$

The $27$ in the denom­i­na­tor can­cels with the $27$ from $3^3$.

Part (b): $(1 - x)^{11}$

Step 1. Here $n = 11$ is odd, so there are $12$ terms and two mid­dle terms: $\displaystyle T_{\frac{12}{2}} = T_6$ (with $r = 5$) and $\displaystyle T_{\frac{14}{2}} = T_7$ (with $r = 6$).

Step 2. The gen­eral term is $T_{r+1} = {}^{11}C_r\,(-x)^r = (-1)^r\,{}^{11}C_r\,x^r$.

Step 3. Find ${}^{11}C_5$. Note that ${}^{11}C_6 = {}^{11}C_5$.

$$\displaystyle {}^{11}C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$$

Step 4. Write both terms.

$$\begin{aligned} T_6 &= (-1)^5 \times 462\,x^5 = -462\,x^5\\ T_7 &= (-1)^6 \times 462\,x^6 = 462\,x^6 \end{aligned}$$

Check­ing the answer#

In (a) there are 3 terms before $T_4$ and 3 after, so it is truly the mid­dle. In (b) there are 5 terms before $T_6$ and 5 after $T_7$.

Answer#

(a) $20x^3y^3$. (b) $-462x^5$ and $462x^6$.

Ques­tion 5: Coef­fi­cient of $a^5b^7$#

The prob­lem#

Find the coef­fi­cient of $a^5b^7$ in $(a - 2b)^{12}$.

Under­stand­ing the prob­lem#

First term $a$, sec­ond term $-2b$, $n = 12$. We want the term con­tain­ing $a^5b^7$ and the num­ber in front of it.

The idea#

In $T_{r+1} = {}^{12}C_r\,a^{12-r}(-2b)^r$ the power of $b$ is $r$. So $r = 7$ (and then $a^{12-7} = a^5$, as required).

Step-by-step solu­tion#

Step 1. Put $r = 7$.

$$T_8 = {}^{12}C_7\,a^{5}\,(-2b)^7 = {}^{12}C_7\,(-2)^7\,a^5b^7$$

Step 2. Find the num­bers.

$$\displaystyle \begin{aligned} {}^{12}C_7 &= {}^{12}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{120} = 792\\ (-2)^7 &= -128 \end{aligned}$$

Step 3. Mul­ti­ply.

$$792 \times (-128) = -101376$$

Check­ing the answer#

An odd power of $-2$ gives a neg­a­tive num­ber, so the coef­fi­cient must be neg­a­tive.

Answer#

The coef­fi­cient of $a^5b^7$ is $-101{,}376$.

Ques­tion 6: A proof about mid­dle terms#

The prob­lem#

Show that the coef­fi­cient of the mid­dle term of $(1 + x)^{2n}$ equals the sum of the coef­fi­cients of the two mid­dle terms of $(1 + x)^{2n-1}$.

Under­stand­ing the prob­lem#

We must write down both sides using bino­mial coef­fi­cients and show they are equal. $(1 + x)^{2n}$ has an even power, so one mid­dle term. $(1 + x)^{2n-1}$ has an odd power, so two mid­dle terms.

The idea#

Find the mid­dle terms using the rules of the les­son, then use Pas­cal's rule ${}^mC_{k-1} + {}^mC_k = {}^{m+1}C_k$.

Step-by-step solu­tion#

Step 1. Mid­dle term of $(1 + x)^{2n}$. The power is $2n$ (even), so the mid­dle term is $\displaystyle T_{\frac{2n}{2}+1} = T_{n+1}$, which has $r = n$.

$$T_{n+1} = {}^{2n}C_n\,x^n, \quad\text{coefficient } {}^{2n}C_n$$

Step 2. Mid­dle terms of $(1 + x)^{2n-1}$. The power is $2n - 1$ (odd), so the mid­dle terms are

$$\displaystyle T_{\frac{(2n-1)+1}{2}} = T_n \quad\text{and}\quad T_{\frac{(2n-1)+3}{2}} = T_{n+1}.$$

They have $r = n - 1$ and $r = n$, so their coef­fi­cients are ${}^{2n-1}C_{n-1}$ and ${}^{2n-1}C_n$.

Step 3. Add these two coef­fi­cients and apply Pas­cal's rule with $m = 2n - 1$, $k = n$.

$${}^{2n-1}C_{n-1} + {}^{2n-1}C_n = {}^{(2n-1)+1}C_n = {}^{2n}C_n$$

Step 4. This is exactly the coef­fi­cient found in Step 1, which proves the state­ment.

Check­ing the answer#

Take $n = 2$: the mid­dle coef­fi­cient of $(1+x)^4$ is ${}^4C_2 = 6$; the mid­dle coef­fi­cients of $(1+x)^3$ are $3$ and $3$, and $3 + 3 = 6$.

Answer#

${}^{2n}C_n = {}^{2n-1}C_{n-1} + {}^{2n-1}C_n$ by Pas­cal's rule, as required.

Ques­tion 7: Equal coef­fi­cients#

The prob­lem#

The coef­fi­cients of $x^2$ and $x^3$ in $(3 + ax)^9$ are equal. Find $a$.

Under­stand­ing the prob­lem#

$a$ is an unknown num­ber (assume $a \ne 0$, oth­er­wise there is no $x$ at all). We write both coef­fi­cients in terms of $a$ and set them equal.

The idea#

The gen­eral term is $T_{r+1} = {}^9C_r\,3^{9-r}(ax)^r = {}^9C_r\,3^{9-r}a^r\,x^r$, so the coef­fi­cient of $x^r$ is ${}^9C_r\,3^{9-r}a^r$.

Step-by-step solu­tion#

Step 1. Coef­fi­cient of $x^2$ ($r = 2$): ${}^9C_2\,3^7\,a^2 = 36 \times 3^7\,a^2$.

Step 2. Coef­fi­cient of $x^3$ ($r = 3$): ${}^9C_3\,3^6\,a^3 = 84 \times 3^6\,a^3$.

Step 3. Set them equal and divide both sides by $3^6 a^2$ (allowed since $a \ne 0$).

$$\displaystyle \begin{aligned} 36 \times 3^7\,a^2 &= 84 \times 3^6\,a^3\\ 36 \times 3 &= 84\,a\\ 108 &= 84\,a\\ a &= \frac{108}{84} = \frac{9}{7} \end{aligned}$$

Check­ing the answer#

With $\displaystyle a = \frac{9}{7}$: $\displaystyle 36 \times 3 \times \left(\frac{9}{7}\right)^2 = \frac{8748}{49}$ and $\displaystyle 84 \times \left(\frac{9}{7}\right)^3 = \frac{8748}{49}$ (both after divid­ing by $3^6$). Equal.

Answer#

$\displaystyle a = \frac{9}{7}$.

Ques­tion 8: Coef­fi­cient in a prod­uct#

The prob­lem#

Find the coef­fi­cient of $x^5$ in $(1 + 2x)^6(1 - x)^7$.

Under­stand­ing the prob­lem#

This is a prod­uct of two bino­mial expan­sions. An $x^5$ term appears when­ever an $x^i$ from the first bracket mul­ti­plies an $x^{5-i}$ from the sec­ond.

The idea#

Coef­fi­cient of $x^i$ in $(1 + 2x)^6$ is ${}^6C_i\,2^i$. Coef­fi­cient of $x^{j}$ in $(1 - x)^7$ is ${}^7C_j\,(-1)^j$. Add the prod­ucts over all pairs with $i + j = 5$, that is $i = 0, 1, \dots, 5$ and $j = 5 - i$.

Step-by-step solu­tion#

Step 1. List the six pairs in a table.

(For exam­ple, ${}^6C_2\,2^2 = 15 \times 4 = 60$ and ${}^7C_3(-1)^3 = -35$.)

Step 2. Add the prod­ucts.

$$-21 + 420 - 2100 + 3360 - 1680 + 192 = 171$$

Check­ing the answer#

Add the pos­i­tives: $420 + 3360 + 192 = 3972$. Add the neg­a­tives: $21 + 2100 + 1680 = 3801$. Then $3972 - 3801 = 171$.

Answer#

The coef­fi­cient of $x^5$ is $171$.

Com­mon mis­take to avoid#

For­get­ting the pairs with $i = 0$ or $i = 5$, where one bracket con­tributes only its con­stant term $1$.

Ques­tion 9: Equal 17th and 18th terms#

The prob­lem#

Find $a$ if the $17$th and $18$th terms of $(2 + a)^{50}$ are equal.

Under­stand­ing the prob­lem#

First term $2$, sec­ond term $a$, $n = 50$. The 17th term has $r = 16$; the 18th term has $r = 17$.

The idea#

Write both terms with $T_{r+1} = {}^{50}C_r\,2^{50-r}a^r$, set them equal, and use ${}^nC_{r}$ ratios to sim­plify.

Step-by-step solu­tion#

Step 1. Write the two terms.

$$T_{17} = {}^{50}C_{16}\,2^{34}a^{16}, \qquad T_{18} = {}^{50}C_{17}\,2^{33}a^{17}$$

Step 2. Set them equal and divide both sides by $2^{33}a^{16}$.

$$\displaystyle {}^{50}C_{16} \times 2 = {}^{50}C_{17}\,a \quad\Rightarrow\quad a = \frac{2\,{}^{50}C_{16}}{{}^{50}C_{17}}$$

Step 3. Use the ratio $\displaystyle \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}$ with $n = 50$, $r = 17$.

$$\displaystyle \frac{{}^{50}C_{17}}{{}^{50}C_{16}} = \frac{50 - 17 + 1}{17} = \frac{34}{17} = 2$$

Step 4. Sub­sti­tute.

$$\displaystyle a = \frac{2}{2} = 1$$

Check­ing the answer#

With $a = 1$ the terms are ${}^{50}C_{16}\,2^{34}$ and ${}^{50}C_{17}\,2^{33} = 2\,{}^{50}C_{16}\,2^{33} = {}^{50}C_{16}\,2^{34}$. Equal.

Answer#

$a = 1$.

Ques­tion 10: Remain­der of $7^{103}$ on divi­sion by 25#

The prob­lem#

Using the the­o­rem, find the remain­der when $7^{103}$ is divided by $25$.

Under­stand­ing the prob­lem#

We want the num­ber $R$ with $0 \le R < 25$ such that $7^{103} = 25q + R$ for some whole num­ber $q$. Com­put­ing $7^{103}$ directly is impos­si­ble by hand, so we use the bino­mial the­o­rem.

The idea#

Notice $7^2 = 49 = 50 - 1$, and $50$ is a mul­ti­ple of $25$. Write $7^{103}$ using $49$, expand $(50 - 1)^{51}$, and see that every term except the last is a mul­ti­ple of $25$.

Step-by-step solu­tion#

Step 1. Rewrite the power.

$$7^{103} = 7 \times 7^{102} = 7 \times (7^2)^{51} = 7 \times 49^{51} = 7\,(50 - 1)^{51}$$

Step 2. Expand $(50 - 1)^{51}$ with the bino­mial the­o­rem.

$$(50 - 1)^{51} = 50^{51} - {}^{51}C_1\,50^{50} + \dots + {}^{51}C_{50}\,50 - 1$$

Every term except the last con­tains at least one fac­tor $50$, so they add up to $50k$ for some whole num­ber $k$.

$$(50 - 1)^{51} = 50k - 1$$

Step 3. Mul­ti­ply by $7$.

$$7^{103} = 7(50k - 1) = 350k - 7 = 25(14k) - 7$$

Step 4. A remain­der can­not be neg­a­tive, so rewrite $-7$ as $-25 + 18$.

$$7^{103} = 25(14k) - 25 + 18 = 25(14k - 1) + 18$$

Since $0 \le 18 < 25$, the remain­der is $18$.

Check­ing the answer#

Small check of the pat­tern: $7^3 = 343 = 13 \times 25 + 18$. Since $7^2 \equiv -1$, $7^4 \equiv 1$, so pow­ers of $7$ repeat every four steps and $7^{103}$ behaves like $7^3$ ($103 - 3 = 100$ is a mul­ti­ple of $4$), giv­ing $18$ again.

Answer#

The remain­der is $18$.

Com­mon mis­take to avoid#

Giv­ing $-7$ as the remain­der. A remain­der on divi­sion by $25$ must lie between $0$ and $24$.