Question

prove that \( 7^n -1\) is divisible by 6 for all n\(\ge \) 1

Answer 1

Using PMI ....

Answer 2

(7)^n -1 = (6+1)^n -1
Now, use binomial coefficient to open (6+1)^n

Answer 3

(6+1)^n = nC0 (6)^0 * 1^n + nC1 * 6^1 + 1^(n-1)+...+nCn * 6^6*1^0
=1 + nC1*6^1*1^(n-1)+...+nCn*6^n *1^0
This gives,
(6+1)^n-1 = nC1*6^1*1^(n-1)+...+nCn*6^n *1^0

Answer 4

Now, u can take 6 common from all the terms of (7)^n-1.
Thus, it is divisible by 6 for all n>=1

Answer 5

just add another way - by induction
check for n =1 we get 6/6
check for n = 2 we get 48/6 = 8
now lets assume that it is correct for n =k
now check for n= k+1
[7^(k+1) + 1] / 6 =
(7*7^k + 1 )/6 =
(6*7^k + 7^k + 1 )/6 =
6*7^k/6 + (7^k + 1)/6
the first term becomes 7^k
and the second by is divisible by our assumption.