Question
using mathematical induction prove that 6^n-1is divisibie by 5, for n>or=0 ?
Answers
oobleck
check for k=0
6^0 -1 = 0 = 0*5
So, P(0)
Assume P(k)
Now consider P(k+1)
6^(k+1) - 1 = 6*6^k - 1 = 6(6^k - 1) + 5
Sine P(k), 6^k - 1 = 5m
So, 6(6^k - 1) + 5 = 6*5m+5 = 5(6m+1)
which is divisible by 5.
So, P(k) => P(k+1)
P(0) => P(1) ... for all k >= 0
6^0 -1 = 0 = 0*5
So, P(0)
Assume P(k)
Now consider P(k+1)
6^(k+1) - 1 = 6*6^k - 1 = 6(6^k - 1) + 5
Sine P(k), 6^k - 1 = 5m
So, 6(6^k - 1) + 5 = 6*5m+5 = 5(6m+1)
which is divisible by 5.
So, P(k) => P(k+1)
P(0) => P(1) ... for all k >= 0