Asked by GEMMA
prove by mathematical induction that 7n+4n+1 is divisible by 6
Answers
Answered by
Steve
As written, it's obviously false. Try n=2
Did you mean 7n^2+4n+1? Nope; false for n=2
Got some error here
Did you mean 7n^2+4n+1? Nope; false for n=2
Got some error here
Answered by
GEMMA
its 7^n+4^n+1
Answered by
GEMMA
Steve
Answered by
Steve
Ah; that's a lot nicer.
it's true for n=1.
So, assume it's true for n=k.
7^(k+1) + 4^(k+1) + 1
= 7*7^k + 4*4^k + 1
= (1+6)*7^k + (1+3)*4^k + 1
= (7^k+4^k+1) + 6*7^k + 3*4^k
obviously, 6*7^k is divisible by 6
4^k is even, so 3*4^k is divisible by 6
So, since we're adding three items which are all multiples of 6, the whole is a multiple of 6.
Thus, the induction step holds.
it's true for n=1.
So, assume it's true for n=k.
7^(k+1) + 4^(k+1) + 1
= 7*7^k + 4*4^k + 1
= (1+6)*7^k + (1+3)*4^k + 1
= (7^k+4^k+1) + 6*7^k + 3*4^k
obviously, 6*7^k is divisible by 6
4^k is even, so 3*4^k is divisible by 6
So, since we're adding three items which are all multiples of 6, the whole is a multiple of 6.
Thus, the induction step holds.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.