a.maths

(3n+1)7^n-1 is divisible for 9 all natural numbers n

For n=1,
(3+1)7^1-1=27 is divisible for 9, n=1 is true

Assume (3n+1)7^n-1 is divisible for 9 for n=k

(3k+1)7^k-1 is divisible for 9
Let (3k+1)7^k-1 = 9A
for n=k+1

[3(k+1)+1)7^(k+1)-1]
=(3k+3+1)7^(k+1)-1
=(3k+1)7^(k+1)+(3)7^(k+1)-1
=(3k+1)7^k x 7 +(3)7^k x 7 -1
=[(3k+1)7^k - 1 ]x 7 + 6 + 3(7^k)7
=9A x 7 + 6+21(7^k)
=63A+6+3[7^(k+1)]

Since the reminder of 3[7^(k+1)] divide by 9 is 6 since 3[7^(k+1)] only have factor of 3 and 7 only. Then 6+3[7^(k+1)] is always divisible by 9.

We can check once again by MI

For k=1, 6+3[7^(1+1)]=153

True for k=m, 6+3[7^(m+1)]=9B, B is an integer.

For k=m+1,
6+3[7^(m+1+1)]
=6+3[7^(m+1)]7
=[6+3[7^(m+1)]]7-6( 6)
=9B(7)-36
=9(7B-4)

It is true 6+3[7^(k+1)] is always divisible by 9.
Then (3n+1)7^n-1 is divisible for 9 all natural numbers n

(3k + 1)．7^k - 1 = 9M,M is integer

when n = k + 1,

(3(k + 1) + 3)．7^(k + 1) - 1

= 7．(7^k)．(3k + 1) + 21．7^k - 1

= 10[(7^k)．(3k + 1) - 1] + 21．7^k + 9 - 3．7^k．(3k + 1)

= 10(9M) + 18．7^k - 9k．7^k + 9

= 9(10M + 2．7^k - k．7^k + 1)