By induction prove for all n e Z Intersection 0 7Intersecti

By induction, prove for all n e Z, Intersection = 0. 7^Intersection - 2^Intersection is divisible by 5

Solution

Let given is x(n)=7ⁿ-2ⁿ   n>0

Need to prove that x(n) is divisible by 5 by using mathematical induction.

x(n)=7ⁿ-2ⁿ

x(1) =7¹-2¹

x(1) = 7-2 = 5

So,x(1) is true and is divisible by 5.

Let x(k) be true, then

Therefore, x(k)= 7^k-2^k   is divisible by 5.

Need to show that x(k+1) is also true and is divisible by 5.

x(k+1) = 7^k+1-2^k+1

Add and subtract 2*7^k

= 7^k+1-2^k+1 +2*7^k - 2*7^k

= 7^k+1 - 2*7^k -2^k+1 +2*7^k

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

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

We already proved that 7^k-2^k is divisible by 5.

So, 7^k (5) is divisible by 5 and 7^k-2^k is divisible by 5.

Therefore, x(k+1) is divisible by 5.

Thus, x(1) is true and x(k+1) is true when x(k) is true.

Therefore, it is proved that 7ⁿ-2ⁿ is divisible by 5 by using induction method.