Pascal's Triangle and the Binomial Theorem, Worked Solutions

Ques­tion 1: Row 7 of Pas­cal's tri­an­gle#

The prob­lem#

Write row $7$ of Pas­cal's tri­an­gle.

Under­stand­ing the prob­lem#

Pas­cal's tri­an­gle starts with row $0$, which is just $1$. Row $n$ lists the num­bers $^nC_0, {}^nC_1, \ldots, {}^nC_n$, so row $7$ has $8$ num­bers. You are asked to write all of them.

The idea#

Use Pas­cal's rule: every entry is the sum of the two entries just above it. The les­son already shows rows $0$ to $5$, so build row $6$ and then row $7$.

Step-by-step solu­tion#

Step 1. Start from row $5$, which the les­son gives.

$$1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1$$

Step 2. Build row $6$. Put $1$ at each end and add neigh­bour­ing pairs of row $5$ in between.

$$\begin{aligned} 1+5 &= 6, & 5+10 &= 15, & 10+10 &= 20, & 10+5 &= 15, & 5+1 &= 6 \end{aligned}$$

So row $6$ is $1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$.

Step 3. Build row $7$ the same way from row $6$.

$$\begin{aligned} 1+6 &= 7, & 6+15 &= 21, & 15+20 &= 35, & 20+15 &= 35, & 15+6 &= 21, & 6+1 &= 7 \end{aligned}$$

Check­ing the answer#

The num­bers in row $n$ add up to $2^n$. Here $1+7+21+35+35+21+7+1 = 128 = 2^7$. Also the row is sym­met­ric, as it must be, and $\displaystyle ^7C_2 = \frac{7 \times 6}{2} = 21$ matches.

Answer#

Row 7: $1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1$.

Com­mon mis­take to avoid#

Do not call $1, 1$ "row 1" and then write the wrong row. Row $n$ always begins $1, n, \ldots$, so row $7$ begins $1, 7$.

Ques­tion 2: Expand­ing three bino­mi­als#

The prob­lem#

Expand: (a) $(x + 2y)^5$; (b) $(3 - x)^4$; (c) $\displaystyle \left(\frac{x}{2} + \frac{2}{x}\right)^4$.

Under­stand­ing the prob­lem#

"Expand" means write the power as a sum of sep­a­rate terms with no brack­ets left. Each is of the form $(a+b)^n$ with a dif­fer­ent $a$, $b$ and $n$.

The idea#

Use the bino­mial the­o­rem

$$\displaystyle (a + b)^n = \sum_{k=0}^{n} {}^nC_k\, a^{n-k} b^k,$$

tak­ing the coef­fi­cients $^nC_k$ from Pas­cal's tri­an­gle. Write each term as (coef­fi­cient) $\times$ (power of $a$) $\times$ (power of $b$), and sim­plify care­fully, because the pow­ers of $2$, $3$ and so on mul­ti­ply into the coef­fi­cients.

Step-by-step solu­tion#

Part (a)

Step 1. Iden­tify the parts: $a = x$, $b = 2y$, $n = 5$. Row $5$ gives coef­fi­cients $1, 5, 10, 10, 5, 1$.

Step 2. Write the six terms, with the power of $x$ falling from $5$ to $0$ and the power of $2y$ ris­ing from $0$ to $5$.

$$(x+2y)^5 = x^5 + 5x^4(2y) + 10x^3(2y)^2 + 10x^2(2y)^3 + 5x(2y)^4 + (2y)^5$$

Step 3. Work out each power of $2y$: $(2y)^2 = 4y^2$, $(2y)^3 = 8y^3$, $(2y)^4 = 16y^4$, $(2y)^5 = 32y^5$. Then mul­ti­ply.

$$\begin{aligned} (x+2y)^5 &= x^5 + 10x^4y + 10\cdot 4\,x^3y^2 + 10 \cdot 8\,x^2y^3 + 5\cdot 16\,xy^4 + 32y^5 \\ &= x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 \end{aligned}$$

Part (b)

Step 1. Here $a = 3$, $b = -x$, $n = 4$; row $4$ is $1, 4, 6, 4, 1$. Because $b$ is neg­a­tive, the signs alter­nate.

Step 2. Write the terms.

$$(3-x)^4 = 3^4 - 4\cdot 3^3 x + 6 \cdot 3^2 x^2 - 4 \cdot 3\, x^3 + x^4$$

Step 3. Eval­u­ate the num­bers: $3^4 = 81$, $4 \cdot 27 = 108$, $6 \cdot 9 = 54$, $4 \cdot 3 = 12$.

$$(3-x)^4 = 81 - 108x + 54x^2 - 12x^3 + x^4$$

Part (c)

Step 1. Here $\displaystyle a = \frac{x}{2}$, $\displaystyle b = \frac{2}{x}$, $n = 4$; coef­fi­cients $1, 4, 6, 4, 1$.

Step 2. Write the terms.

$$\displaystyle \left(\frac{x}{2} + \frac{2}{x}\right)^4 = \left(\frac{x}{2}\right)^4 + 4\left(\frac{x}{2}\right)^3\left(\frac{2}{x}\right) + 6\left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right)^2 + 4\left(\frac{x}{2}\right)\left(\frac{2}{x}\right)^3 + \left(\frac{2}{x}\right)^4$$

Step 3. Sim­plify each term sep­a­rately.

$$\displaystyle \begin{aligned} \left(\frac{x}{2}\right)^4 &= \frac{x^4}{16} \\ 4 \cdot \frac{x^3}{8} \cdot \frac{2}{x} &= \frac{8x^3}{8x} = x^2 \\ 6 \cdot \frac{x^2}{4} \cdot \frac{4}{x^2} &= 6 \\ 4 \cdot \frac{x}{2} \cdot \frac{8}{x^3} &= \frac{32x}{2x^3} = \frac{16}{x^2} \\ \left(\frac{2}{x}\right)^4 &= \frac{16}{x^4} \end{aligned}$$

Step 4. Put them together.

$$\displaystyle \left(\frac{x}{2} + \frac{2}{x}\right)^4 = \frac{x^4}{16} + x^2 + 6 + \frac{16}{x^2} + \frac{16}{x^4}$$

Check­ing the answer#

Put $x = y = 1$ in (a): the left side is $3^5 = 243$ and the right side is $1+10+40+80+80+32 = 243$. Put $x = 1$ in (b): $2^4 = 16$ and $81 - 108 + 54 - 12 + 1 = 16$. Put $x = 2$ in (c): $(1+1)^4 = 16$ and $1 + 4 + 6 + 4 + 1 = 16$. All three agree.

Answer#

(a) $x^5 + 10x^4 y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5$; (b) $81 - 108x + 54x^2 - 12x^3 + x^4$; (c) $\displaystyle \frac{x^4}{16} + x^2 + 6 + \frac{16}{x^2} + \frac{16}{x^4}$.

Com­mon mis­take to avoid#

Raise the whole of $b$ to the power, not just the let­ter: $(2y)^3 = 8y^3$, not $2y^3$.

Ques­tion 3: Com­put­ing pow­ers with the the­o­rem#

The prob­lem#

Com­pute using the bino­mial the­o­rem: (a) $(102)^5$; (b) $(98)^4$; (c) $(1.1)^5$.

Under­stand­ing the prob­lem#

Mul­ti­ply­ing these num­bers out directly is long and error-prone. You are asked to write each base as a round num­ber plus or minus a small num­ber, and then expand.

The idea#

Write $102 = 100 + 2$, $98 = 100 - 2$ and $1.1 = 1 + 0.1$. Pow­ers of $100$ and of $0.1$ are easy, so each term of the expan­sion is easy to com­pute, exactly as in Exam­ple 3 of the les­son.

Step-by-step solu­tion#

Part (a)

Step 1. Write $(102)^5 = (100 + 2)^5$ and use row $5$: $1, 5, 10, 10, 5, 1$.

$$(100+2)^5 = 100^5 + 5\cdot 100^4 \cdot 2 + 10 \cdot 100^3 \cdot 2^2 + 10\cdot 100^2 \cdot 2^3 + 5 \cdot 100 \cdot 2^4 + 2^5$$

Step 2. Eval­u­ate each term.

$$\begin{aligned} 100^5 &= 10{,}000{,}000{,}000 \\ 5 \cdot 100^4 \cdot 2 &= 10 \cdot 10^8 = 1{,}000{,}000{,}000 \\ 10 \cdot 10^6 \cdot 4 &= 40{,}000{,}000 \\ 10 \cdot 10^4 \cdot 8 &= 800{,}000 \\ 5 \cdot 100 \cdot 16 &= 8{,}000 \\ 2^5 &= 32 \end{aligned}$$

Step 3. Add.

$$10{,}000{,}000{,}000 + 1{,}000{,}000{,}000 + 40{,}000{,}000 + 800{,}000 + 8{,}000 + 32 = 11{,}040{,}808{,}032$$

Part (b)

Step 1. Write $(98)^4 = (100 - 2)^4$; row $4$ is $1, 4, 6, 4, 1$, signs alter­nate.

$$(100-2)^4 = 100^4 - 4 \cdot 100^3 \cdot 2 + 6 \cdot 100^2 \cdot 4 - 4 \cdot 100 \cdot 8 + 16$$

Step 2. Eval­u­ate each term.

$$100^4 = 100{,}000{,}000,\quad 8{,}000{,}000,\quad 240{,}000,\quad 3{,}200,\quad 16$$

Step 3. Com­bine with the signs.

$$\begin{aligned} &100{,}000{,}000 - 8{,}000{,}000 + 240{,}000 - 3{,}200 + 16 \\ &= 92{,}000{,}000 + 240{,}000 - 3{,}200 + 16 \\ &= 92{,}240{,}000 - 3{,}200 + 16 = 92{,}236{,}816 \end{aligned}$$

Part (c)

Step 1. Write $(1.1)^5 = (1 + 0.1)^5$.

$$(1+0.1)^5 = 1 + 5(0.1) + 10(0.1)^2 + 10(0.1)^3 + 5(0.1)^4 + (0.1)^5$$

Step 2. Eval­u­ate each term.

$$1 + 0.5 + 0.1 + 0.01 + 0.0005 + 0.00001$$

Step 3. Add: $1.61051$.

Check­ing the answer#

Esti­mates: $102^5$ is a lit­tle more than $100^5 = 10^{10}$, and $1.1 \times 10^{10}$ is sen­si­ble. $98^4$ is a lit­tle less than $10^8$, as found. $1.1^2 = 1.21$, $1.1^4 = 1.4641$, $1.4641 \times 1.1 = 1.61051$, which matches.

Answer#

(a) $11{,}040{,}808{,}032$; (b) $92{,}236{,}816$; (c) $1.61051$.

Ques­tion 4: Com­par­ing $(1.01)^{100}$ with 2#

The prob­lem#

Which is larger: $(1.01)^{100}$ or $2$?

Under­stand­ing the prob­lem#

You can­not eas­ily mul­ti­ply $1.01$ by itself $100$ times. You only need to decide which num­ber is big­ger, not find $(1.01)^{100}$ exactly.

The idea#

Expand $(1 + 0.01)^{100}$ and look at just the first two terms. Every term of the expan­sion is pos­i­tive, so the whole sum is larger than any part of it. This is the method of Exam­ple 4 in the les­son.

Step-by-step solu­tion#

Step 1. Write the expan­sion.

$$(1 + 0.01)^{100} = 1 + {}^{100}C_1 (0.01) + {}^{100}C_2 (0.01)^2 + \cdots$$

Step 2. The first two terms add to exactly $2$.

$$1 + 100 \times 0.01 = 1 + 1 = 2$$

Step 3. All the remain­ing terms are pos­i­tive (pos­i­tive coef­fi­cients times pow­ers of a pos­i­tive num­ber). The third term alone is $\displaystyle \frac{100 \times 99}{2}(0.0001) = 0.495$. So

$$(1.01)^{100} = 2 + 0.495 + (\text{more positive terms}) > 2.$$

Check­ing the answer#

A cal­cu­la­tor gives $(1.01)^{100} \approx 2.705$, which is indeed larger than $2$.

Answer#

$(1.01)^{100}$ is larger than $2$.

Ques­tion 5: Sum of two sixth pow­ers#

The prob­lem#

Find $(x + 1)^6 + (x - 1)^6$ and hence find $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$.

Under­stand­ing the prob­lem#

First find a sim­pli­fied for­mula in $x$, then put $x = \sqrt{2}$ in that for­mula. The word "hence" tells you to use the first result for the sec­ond part.

The idea#

Expand both with row $6$: $1, 6, 15, 20, 15, 6, 1$. In $(x-1)^6$ the terms with odd pow­ers of $1$ change sign, so when you add, the odd terms can­cel and the even terms dou­ble (com­pare Exam­ple 5).

Step-by-step solu­tion#

Step 1. Expand $(x+1)^6$.

$$(x+1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$$

Step 2. Expand $(x-1)^6$; the signs alter­nate.

$$(x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$$

Step 3. Add. The $x^5$, $x^3$ and $x$ terms can­cel; the oth­ers dou­ble.

$$(x+1)^6 + (x-1)^6 = 2x^6 + 30x^4 + 30x^2 + 2 = 2(x^6 + 15x^4 + 15x^2 + 1)$$

Step 4. Put $x = \sqrt{2}$. Then $x^2 = 2$, $x^4 = 4$, $x^6 = 8$.

$$\begin{aligned} (\sqrt{2}+1)^6 + (\sqrt{2}-1)^6 &= 2(8 + 15 \cdot 4 + 15 \cdot 2 + 1) \\ &= 2(8 + 60 + 30 + 1) = 2 \times 99 = 198 \end{aligned}$$

Check­ing the answer#

$\sqrt{2} + 1 \approx 2.414$, and $2.414^6 \approx 197.99$, while $(\sqrt{2} - 1)^6 \approx 0.005$. Their sum is about $198$. It is also sen­si­ble that the answer is a whole num­ber, because all the $\sqrt{2}$ terms came from odd pow­ers and can­celled.

Answer#

$(x+1)^6 + (x-1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)$, and $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 198$.

Ques­tion 6: A divis­i­bil­ity proof#

The prob­lem#

Show that $9^{n+1} - 8n - 9$ is divis­i­ble by $64$ for every pos­i­tive inte­ger $n$.

Under­stand­ing the prob­lem#

You must prove that, what­ever pos­i­tive inte­ger $n$ is, the num­ber $9^{n+1} - 8n - 9$ equals $64$ times a whole num­ber. Check­ing a few val­ues is not a proof; you need an argu­ment that works for every $n$.

The idea#

Write $9 = 1 + 8$ and expand $(1 + 8)^{n+1}$. The first two terms will can­cel with $-8n - 9$, and every later term con­tains $8^2 = 64$. This is the method of Exam­ple 6.

Step-by-step solu­tion#

Step 1. Expand with the bino­mial the­o­rem, using $m = n+1$ as the power.

$$9^{n+1} = (1+8)^{n+1} = 1 + {}^{n+1}C_1 \cdot 8 + {}^{n+1}C_2 \cdot 8^2 + {}^{n+1}C_3 \cdot 8^3 + \cdots + 8^{n+1}$$

Step 2. Sim­plify the first two terms: $^{n+1}C_1 = n + 1$, so $1 + 8(n+1) = 8n + 9$.

Step 3. Take out $8^2 = 64$ from every remain­ing term.

$$9^{n+1} = 8n + 9 + 64\left[{}^{n+1}C_2 + {}^{n+1}C_3 \cdot 8 + \cdots + 8^{n-1}\right]$$

Step 4. Sub­tract $8n + 9$ from both sides.

$$9^{n+1} - 8n - 9 = 64\left[{}^{n+1}C_2 + {}^{n+1}C_3 \cdot 8 + \cdots + 8^{n-1}\right]$$

Step 5. The bracket is a sum of whole num­bers (bino­mial coef­fi­cients times pow­ers of $8$), so it is a whole num­ber. Hence the left side is a mul­ti­ple of $64$.

Check­ing the answer#

$n = 1$: $81 - 8 - 9 = 64$. $n = 2$: $729 - 16 - 9 = 704 = 64 \times 11$. Both are mul­ti­ples of $64$.

Answer#

$9^{n+1} - 8n - 9 = 64 \times (\text{a whole number})$, so it is divis­i­ble by $64$ for every pos­i­tive inte­ger $n$.

Ques­tion 7: Sums of bino­mial coef­fi­cients#

The prob­lem#

Find the sum $^{10}C_0 + {}^{10}C_1 + \cdots + {}^{10}C_{10}$ and the sum of the even-posi­tion coef­fi­cients $^{10}C_0 + {}^{10}C_2 + \cdots + {}^{10}C_{10}$.

Under­stand­ing the prob­lem#

The first sum is all eleven coef­fi­cients of $(a+b)^{10}$. The sec­ond takes only $^{10}C_0, {}^{10}C_2, {}^{10}C_4, \ldots, {}^{10}C_{10}$.

The idea#

The les­son shows two facts: putting $a = b = 1$ gives $\displaystyle \sum {}^nC_k = 2^n$; putting $a = 1$, $b = -1$ gives $^nC_0 - {}^nC_1 + {}^nC_2 - \cdots = 0$. Adding these two facts iso­lates the even-posi­tion terms.

Step-by-step solu­tion#

Step 1. Put $a = b = 1$ in $(a+b)^{10}$.

$$^{10}C_0 + {}^{10}C_1 + \cdots + {}^{10}C_{10} = 2^{10} = 1024$$

Step 2. Put $a = 1$, $b = -1$.

$$^{10}C_0 - {}^{10}C_1 + {}^{10}C_2 - \cdots + {}^{10}C_{10} = 0$$

Step 3. Call the even-posi­tion sum $E$ and the odd-posi­tion sum $O$. The two lines say $E + O = 1024$ and $E - O = 0$.

Step 4. Add the two equa­tions: $2E = 1024$, so $E = 512$.

Check­ing the answer#

Row $10$ is $1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1$. Even posi­tions: $1 + 45 + 210 + 210 + 45 + 1 = 512$. Cor­rect.

Answer#

The full sum is $1024$ and the even-posi­tion sum is $512$.

Ques­tion 8: Expand­ing a tri­no­mial#

The prob­lem#

Expand $(1 + x + x^2)^3$ using $(1 + (x + x^2))^3$.

Under­stand­ing the prob­lem#

This has three terms inside the bracket, but the bino­mial the­o­rem needs two. The hint groups $x + x^2$ as a sin­gle term $b$.

The idea#

Use $(1 + b)^3 = 1 + 3b + 3b^2 + b^3$ with $b = x + x^2$, then expand each power of $b$ (each is again a bino­mial) and col­lect like terms.

Step-by-step solu­tion#

Step 1. Apply the the­o­rem with $b = x + x^2$.

$$(1 + b)^3 = 1 + 3b + 3b^2 + b^3$$

Step 2. Find the pow­ers of $b$.

$$\begin{aligned} b &= x + x^2 \\ b^2 &= x^2 + 2x^3 + x^4 \\ b^3 &= x^3 + 3x^4 + 3x^5 + x^6 \end{aligned}$$

Step 3. Mul­ti­ply by the coef­fi­cients.

$$\begin{aligned} 3b &= 3x + 3x^2 \\ 3b^2 &= 3x^2 + 6x^3 + 3x^4 \\ b^3 &= x^3 + 3x^4 + 3x^5 + x^6 \end{aligned}$$

Step 4. Add every­thing, col­lect­ing each power of $x$.

$$\begin{aligned} x^0 &: 1 \\ x^1 &: 3 \\ x^2 &: 3 + 3 = 6 \\ x^3 &: 6 + 1 = 7 \\ x^4 &: 3 + 3 = 6 \\ x^5 &: 3 \\ x^6 &: 1 \end{aligned}$$

Check­ing the answer#

Put $x = 1$: the left side is $3^3 = 27$, and $1 + 3 + 6 + 7 + 6 + 3 + 1 = 27$. The coef­fi­cients are also sym­met­ric, as expected from the sym­met­ric bracket $1 + x + x^2$.

Answer#

$(1 + x + x^2)^3 = 1 + 3x + 6x^2 + 7x^3 + 6x^4 + 3x^5 + x^6$.