Exercise 7.2 — Question 15

Problem Statement

Using the Binomial Theorem, show that $3^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ .

Solution

We need to show that $8 ∣ (3^{2 n} - 1)$ for all $n \in Z^{+}$ .

Step 1: Rewrite the base.

$3^{2 n} = (3^{2})^{n} = 9^{n}$

So we need to show $9^{n} - 1$ is divisible by $8$ .

Step 2: Express $9$ in a form suitable for the Binomial Theorem.

$9^{n} = (1 + 8)^{n}$

Step 3: Expand using the Binomial Theorem.

$(1 + 8)^{n} = (0 n) 1^{n} + (1 n) 1^{n - 1} (8) + (2 n) 1^{n - 2} (8)^{2} + \dots + (n n) (8)^{n}$

$= 1 + 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots + 8^{n}$

Step 4: Subtract 1 and factor out 8.

$9^{n} - 1 = 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots + 8^{n}$

$= 8 [n + (2 n) 8 + (3 n) 8^{2} + \dots + 8^{n - 1}]$

Since every term inside the bracket is an integer, $9^{n} - 1$ is divisible by $8$ .

Therefore, $3^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ . $■$

Key Technique: Divisibility via Binomial Theorem

To prove $N^{n} - 1$ is divisible by $(N - 1)$ , write $N = 1 + (N - 1)$ and expand:

$(1 + k)^{n} = 1 + nk + (2 n) k^{2} + \dots$

Subtracting 1 leaves every term divisible by $k = N - 1$ .

This technique generalises: to test divisibility by $d$ , write the base as $1 + d$ or $m d + r$ and expand.

Exercise 7.2 — Question 15

Problem Statement

Using the Binomial Theorem, show that $3^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ .

Solution

We need to show that $8 ∣ (3^{2 n} - 1)$ for all $n \in Z^{+}$ .

Step 1: Rewrite the base.

$3^{2 n} = (3^{2})^{n} = 9^{n}$

So we need to show $9^{n} - 1$ is divisible by $8$ .

Step 2: Express $9$ in a form suitable for the Binomial Theorem.

$9^{n} = (1 + 8)^{n}$

Step 3: Expand using the Binomial Theorem.

$(1 + 8)^{n} = (0 n) 1^{n} + (1 n) 1^{n - 1} (8) + (2 n) 1^{n - 2} (8)^{2} + \dots + (n n) (8)^{n}$

$= 1 + 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots + 8^{n}$

Step 4: Subtract 1 and factor out 8.

$9^{n} - 1 = 8 n + (2 n) 8^{2} + (3 n) 8^{3} + \dots + 8^{n}$

$= 8 [n + (2 n) 8 + (3 n) 8^{2} + \dots + 8^{n - 1}]$

Since every term inside the bracket is an integer, $9^{n} - 1$ is divisible by $8$ .

Therefore, $3^{2 n} - 1$ is divisible by $8$ for all positive integers $n$ . $■$

Key Technique: Divisibility via Binomial Theorem

To prove $N^{n} - 1$ is divisible by $(N - 1)$ , write $N = 1 + (N - 1)$ and expand:

$(1 + k)^{n} = 1 + nk + (2 n) k^{2} + \dots$

Subtracting 1 leaves every term divisible by $k = N - 1$ .

This technique generalises: to test divisibility by $d$ , write the base as $1 + d$ or $m d + r$ and expand.

7.2 Q-15

Exercise 7.2 — Question 15

Problem Statement

Solution

Key Technique: Divisibility via Binomial Theorem

7.2 Q-15

Exercise 7.2 — Question 15

Problem Statement

Solution

Key Technique: Divisibility via Binomial Theorem