Question

Prove by method of induction, for all n ∈ N
$(2^{3n}-1)$ is divisible by 7.

Answer 1

Solution :-

Let P(n): 23n - 1 is divisible by 7

Now , P( 1): 23 - 1 = 7, which is divisible by 7 .

Hence, P(l) is true .

Let us assume that P(n) is true for some natural number n = k.

P(k): 23k – 1 is divisible by 7.

or  23k -1 = 7m, m∈ N  

Now, we have to prove that P(k + 1) is true.

P(k+ 1): 23(k+1)– 1

= 23k.23 -1

= 8(7 m + 1) – 1

= 56 m + 7

= 7(8m + 1), which is divisible by 7.



So , by the principle of mathematical induction P(n) is true for all natural numbers n .

Hence , Proved .

Answer 2

Answer:

Method 1

2^(3n) - 1

= 2^(3n) - 1^n

= 8^n - 1^n

= (8-1)(8^(n-1) + .....+ 1)

= 7 x Some Integer

This is divisible by 7

Method 2

If you are not comfortable with the expansion of a^n - b^n, you may also use mathematical induction to prove that 2^(3n) - 1 is divisible by 7

for n = 0 expression evaluates to 0 --> divisible by 7

if the expression is divisible by 7 for n = k (an integer)

2(3k) - 1 = 7 * C ; where C is an integer constant

Writing 2^3 as 8 for simplicity,

8^k -1 = 7C

Now multiply both sides by 8.

8^(k+1) - 8 = 7 * 8 * C

8^(k+1) - 1 = 7 * 8 * C + 7 = 7 (8C + 1)

8^(k+1) - 1 = 7 * (An Integer)

Writing 8 as 2^3 again,

2^[3(k+1)] -1 = 7 * An integer --> R.H.S is divisible by 7

Therefore if the statement is true for n = k, it is true for n = k+1

Thus by mathematical induction 2^(3n) - 1 is divisible by 7 for all non-negative integer values of n.