Exercise 7.4 — Question 3

Topic: Applications of the Binomial Theorem

This exercise focuses on applying the Binomial Theorem to:

Expand and simplify binomial expressions
Find remainders when large powers are divided by a number
Determine the last digit of a large power
Test divisibility using the Binomial Theorem

Key Formula: Binomial Theorem

For any positive integer $n$ :

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

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

Technique 1: Finding the Remainder

To find the remainder when a large power is divided by a number, rewrite the base as a sum involving that divisor.

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

Solution: $7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} \cdot 1^{100 - r}$ $= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

Every term except the first ( $r = 0$ ) contains a factor of $6$ . Therefore: $7^{100} = 6 k + 1 for some integer k$

Remainder = 1

Technique 2: Finding the Last Digit

The last (units) digit of a number equals its remainder when divided by $10$ . Rewrite the base to isolate a multiple of $10$ .

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

Solution: $3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$ $= \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

Every term with $r \geq 1$ is divisible by $10$ . The $r = 0$ term is $(- 1)^{50} = 1$ .

So $3^{100} = 10 m + 1$ for some integer $m$ .

Last digit = 1

Technique 3: Testing Divisibility

Rewrite the expression so that the Binomial Theorem isolates a multiple of the divisor.

Example: Show that $6^{n} - 1$ is divisible by $5$ for all positive integers $n$ .

Solution: $6^{n} = (5 + 1)^{n} = \sum_{r = 0}^{n} (r n) 5^{r} = 1 + (1 n) \cdot 5 + (2 n) \cdot 5^{2} + \dots + 5^{n}$

Therefore: $6^{n} - 1 = 5 [n + (2 n) \cdot 5 + \dots + 5^{n - 1}]$

This is clearly divisible by $5$ . $■$

Technique 4: Expanding and Simplifying

Example: Expand $(x + 2)^{4}$ using the Binomial Theorem.

$(x + 2)^{4} = (0 4) x^{4} + (1 4) x^{3} (2) + (2 4) x^{2} (4) + (3 4) x (8) + (4 4) (16)$ $= x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16$

Exercise 7.4 — Question 3

Topic: Applications of the Binomial Theorem

This exercise focuses on applying the Binomial Theorem to:

Expand and simplify binomial expressions
Find remainders when large powers are divided by a number
Determine the last digit of a large power
Test divisibility using the Binomial Theorem

Key Formula: Binomial Theorem

For any positive integer $n$ :

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

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

Technique 1: Finding the Remainder

To find the remainder when a large power is divided by a number, rewrite the base as a sum involving that divisor.

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

Solution: $7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} \cdot 1^{100 - r}$ $= 1 + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots + 6^{100}$

Every term except the first ( $r = 0$ ) contains a factor of $6$ . Therefore: $7^{100} = 6 k + 1 for some integer k$

Remainder = 1

Technique 2: Finding the Last Digit

The last (units) digit of a number equals its remainder when divided by $10$ . Rewrite the base to isolate a multiple of $10$ .

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

Solution: $3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50}$ $= \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

Every term with $r \geq 1$ is divisible by $10$ . The $r = 0$ term is $(- 1)^{50} = 1$ .

So $3^{100} = 10 m + 1$ for some integer $m$ .

Last digit = 1

Technique 3: Testing Divisibility

Rewrite the expression so that the Binomial Theorem isolates a multiple of the divisor.

Example: Show that $6^{n} - 1$ is divisible by $5$ for all positive integers $n$ .

Solution: $6^{n} = (5 + 1)^{n} = \sum_{r = 0}^{n} (r n) 5^{r} = 1 + (1 n) \cdot 5 + (2 n) \cdot 5^{2} + \dots + 5^{n}$

Therefore: $6^{n} - 1 = 5 [n + (2 n) \cdot 5 + \dots + 5^{n - 1}]$

This is clearly divisible by $5$ . $■$

Technique 4: Expanding and Simplifying

Example: Expand $(x + 2)^{4}$ using the Binomial Theorem.

$(x + 2)^{4} = (0 4) x^{4} + (1 4) x^{3} (2) + (2 4) x^{2} (4) + (3 4) x (8) + (4 4) (16)$ $= x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16$

7.4 Q-3

Exercise 7.4 — Question 3

Topic: Applications of the Binomial Theorem

Key Formula: Binomial Theorem

Technique 1: Finding the Remainder

Technique 2: Finding the Last Digit

Technique 3: Testing Divisibility

Technique 4: Expanding and Simplifying

7.4 Q-3

Exercise 7.4 — Question 3

Topic: Applications of the Binomial Theorem

Key Formula: Binomial Theorem

Technique 1: Finding the Remainder

Technique 2: Finding the Last Digit

Technique 3: Testing Divisibility

Technique 4: Expanding and Simplifying