Exercise 7.4 — Question 4

Binomial Theorem: Applications

This question set covers advanced applications of the Binomial Theorem including approximations, remainders, last digits, and divisibility tests.

Key Formula

The Binomial Theorem states that for any positive integer $n$ :

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

where $(r n) = \frac{n !}{r ! ( n - r )!}$ .

Application 1: Expansion and Simplification (SLO M-11-A-34, M-11-A-37)

To expand $(a + b)^{n}$ , apply the Binomial Theorem term by term. Each term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Example: Expand $(2 x - 3)^{4}$ .

$(2 x - 3)^{4} = \sum_{r = 0}^{4} (r 4) (2 x)^{4 - r} (- 3)^{r}$

$= 16 x^{4} - 4 \cdot 8 x^{3} \cdot 3 + 6 \cdot 4 x^{2} \cdot 9 - 4 \cdot 2 x \cdot 27 + 81$

$= 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$

Application 2: Approximate Values (SLO M-11-A-38)

For small $x$ , write the expression in the form $(1 + x)^{n}$ and expand, keeping only the first few terms.

Example: Find an approximate value of $(1.02)^{10}$ .

Write $1.02 = 1 + 0.02$ :

$(1.02)^{10} \approx 1 + 10 (0.02) + \frac{10 \cdot 9}{2} (0.02)^{2} + \dots$

$\approx 1 + 0.2 + 0.018 + \dots \approx 1.218$

Application 3: Finding Remainders (SLO M-11-A-39)

Express the base as $(divisor \pm 1)$ or a convenient form, then expand using the Binomial Theorem. All terms except the constant will be divisible by the divisor.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

Write $7 = 6 + 1$ :

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r}$

$= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots$

Every term except $1$ is divisible by $6$ , so:

$7^{100} = 6 k + 1 for some integer k$

Remainder = 1

Application 4: Last Digit, Divisibility, and Comparing Large Numbers (SLO M-11-A-40)

Last Digit

Express the base so that the last digit pattern is clear. Use the Binomial Theorem to isolate the units digit.

Example: Find the last digit of $3^{100}$ .

Note $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Write $81 = 80 + 1$ :

$8 1^{25} = (80 + 1)^{25} = 1 + 25 \cdot 80 + \dots$

All terms except $1$ end in $0$ , so the last digit of $3^{100}$ is 1.

Divisibility Test

Example: Show $6^{n} - 1$ is divisible by $5$ .

Write $6 = 5 + 1$ :

$6^{n} = (5 + 1)^{n} = 1 + 5 n + (2 n) 5^{2} + \dots$

$\Rightarrow 6^{n} - 1 = 5 n + (2 n) 5^{2} + \dots = 5 (n + (2 n) 5 + \dots)$

Hence $6^{n} - 1$ is divisible by $5$ . ✓

Comparing Large Numbers

To compare $e . g ., 2^{50}$ vs $3^{33}$ , express both in a common base form using the Binomial Theorem and compare the leading terms.

Exercise 7.4 — Question 4

Binomial Theorem: Applications

This question set covers advanced applications of the Binomial Theorem including approximations, remainders, last digits, and divisibility tests.

Key Formula

The Binomial Theorem states that for any positive integer $n$ :

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

where $(r n) = \frac{n !}{r ! ( n - r )!}$ .

Application 1: Expansion and Simplification (SLO M-11-A-34, M-11-A-37)

To expand $(a + b)^{n}$ , apply the Binomial Theorem term by term. Each term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}$

Example: Expand $(2 x - 3)^{4}$ .

$(2 x - 3)^{4} = \sum_{r = 0}^{4} (r 4) (2 x)^{4 - r} (- 3)^{r}$

$= 16 x^{4} - 4 \cdot 8 x^{3} \cdot 3 + 6 \cdot 4 x^{2} \cdot 9 - 4 \cdot 2 x \cdot 27 + 81$

$= 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$

Application 2: Approximate Values (SLO M-11-A-38)

For small $x$ , write the expression in the form $(1 + x)^{n}$ and expand, keeping only the first few terms.

Example: Find an approximate value of $(1.02)^{10}$ .

Write $1.02 = 1 + 0.02$ :

$(1.02)^{10} \approx 1 + 10 (0.02) + \frac{10 \cdot 9}{2} (0.02)^{2} + \dots$

$\approx 1 + 0.2 + 0.018 + \dots \approx 1.218$

Application 3: Finding Remainders (SLO M-11-A-39)

Express the base as $(divisor \pm 1)$ or a convenient form, then expand using the Binomial Theorem. All terms except the constant will be divisible by the divisor.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

Write $7 = 6 + 1$ :

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r}$

$= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots$

Every term except $1$ is divisible by $6$ , so:

$7^{100} = 6 k + 1 for some integer k$

Remainder = 1

Application 4: Last Digit, Divisibility, and Comparing Large Numbers (SLO M-11-A-40)

Last Digit

Express the base so that the last digit pattern is clear. Use the Binomial Theorem to isolate the units digit.

Example: Find the last digit of $3^{100}$ .

Note $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Write $81 = 80 + 1$ :

$8 1^{25} = (80 + 1)^{25} = 1 + 25 \cdot 80 + \dots$

All terms except $1$ end in $0$ , so the last digit of $3^{100}$ is 1.

Divisibility Test

Example: Show $6^{n} - 1$ is divisible by $5$ .

Write $6 = 5 + 1$ :

$6^{n} = (5 + 1)^{n} = 1 + 5 n + (2 n) 5^{2} + \dots$

$\Rightarrow 6^{n} - 1 = 5 n + (2 n) 5^{2} + \dots = 5 (n + (2 n) 5 + \dots)$

Hence $6^{n} - 1$ is divisible by $5$ . ✓

Comparing Large Numbers

To compare $e . g ., 2^{50}$ vs $3^{33}$ , express both in a common base form using the Binomial Theorem and compare the leading terms.

7.4 Q-4

Exercise 7.4 — Question 4

Binomial Theorem: Applications

Key Formula

Application 1: Expansion and Simplification (SLO M-11-A-34, M-11-A-37)

Application 2: Approximate Values (SLO M-11-A-38)

Application 3: Finding Remainders (SLO M-11-A-39)

Application 4: Last Digit, Divisibility, and Comparing Large Numbers (SLO M-11-A-40)

Last Digit

Divisibility Test

Comparing Large Numbers

7.4 Q-4

Exercise 7.4 — Question 4

Binomial Theorem: Applications

Key Formula

Application 1: Expansion and Simplification (SLO M-11-A-34, M-11-A-37)

Application 2: Approximate Values (SLO M-11-A-38)

Application 3: Finding Remainders (SLO M-11-A-39)

Application 4: Last Digit, Divisibility, and Comparing Large Numbers (SLO M-11-A-40)

Last Digit

Divisibility Test

Comparing Large Numbers