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
prove by mathematical induction that 7n+4n+1 is divisible by 6
4 answers
its 7^n+4^n+1
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.
4 answers