Question

prove (8^k) - (2^k) is divisible by 6 for n belonging to 0, 1, 2, 3, n,..n+1

Answer 1

\[8^{k} = 2^{3k}\]
\[2^{3k} -2^{k} = 2^{k}(4^{k}-1)\]

Answer 2

one of the proof is as follows
(8^k) - (2^k)
=(8- 2) ((8)^(k-1) +(8)^(k-1)(2)+(8)^(k-2)(2)^2 ...................(8)(2)^(k-2)+(2)^(k-1))
=6 ((8)^(k-1) +(8)^(k-1)(2)+(8)^(k-2)(2)^2 ...................(8)(2)^(k-2)+(2)^(k-1))

Answer 3

1. bASE STEP
n=0
1-1=0

2. kth step
assume 8^k - 2^k is div by 6

3. k+1 step
8^k +1 - (2^(k+1)) = 8(8^k -2^k) + 8 * 2^k + (2^(k+1))=
= 8(8^k -2^k) + 8 * 2^k + 2^k * 2= 8(8^k - 2^k) + 6 *2^k
since (8^k - 2^k) divisble by 6 by assumption in kth step and 6*2^k is divisible by 6, it follows that the sum is divisible by 6

Answer 4

using binomials (8^k) - (2^k) =((6+2))^k - (2^k)

expand the first expression it is found that 6 will be one of the factor...