Question

Prove that for any natural number n, 7^n-2^n is divisible by 5

Answer 1

Answer:

Let P(n): 7n – 2n is divisible by 5, for any natural number n.

Now, P(l) = 71-21 = 5, which is divisible by 5.

Hence, P(l) is true.

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

.’. P(k) = 7k -2k is divisible by 5

or 7k – 2k = 5m, m∈ N (i)

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

P(k+ 1): 7k+1 -2k+1

= 7k-7-2k-2

= (5 + 2)7k -2k-2

= 5.7k + 2.7k -2-2k

= 5.7k + 2(7k – 2k)

= 5 • 7k + 2(5 m) (using (i))

= 5(7k + 2m), which divisible by 5.

Thus, P(k + 1) is true whenever P(k) is true.

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

hope it helped u o==[]::::::::::::::::>\(-ㅂ-)/ ♥ ♥ ♥

Answer 2

Answer:

Let P(n): 7n – 2n is divisible by 5, for any natural number n.

Now, P(l) = 71-21 = 5, which is divisible by 5.

Hence, P(l) is true.

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

.’. P(k) = 7k -2k is divisible by 5

or 7k – 2k = 5m, m∈ N (i)

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

P(k+ 1): 7k+1 -2k+1

= 7k-7-2k-2

= (5 + 2)7k -2k-2

= 5.7k + 2.7k -2-2k

= 5.7k + 2(7k – 2k)

= 5 • 7k + 2(5 m) (using (i))

= 5(7k + 2m), which divisible by 5.

Thus, P(k + 1) is true whenever P(k) is true.

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

hope it helps u ✌