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Xyius
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I need to prove by mathematical induction that 6 divides (k^2+5)k for all k>=1. It works for 1 but I am having trouble obtaining the k+1'st term from the k'th term. Can anyone help? Thanks!
alphachapmtl said:if k=0 mod 6, k^2+5 = 5 mod 6, (k^2+5)k = 0 mod 6
if k=1 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6
if k=2 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=3 mod 6, k^2+5 = 2 mod 6, (k^2+5)k = 0 mod 6
if k=4 mod 6, k^2+5 = 3 mod 6, (k^2+5)k = 0 mod 6
if k=5 mod 6, k^2+5 = 0 mod 6, (k^2+5)k = 0 mod 6
See Cubzar's posr. The other proofs don't seem to be using induction. You showed it for k = 1. I think that to prove by induction you, you would have to write k+1 where ever k appears in the equation and expand the equation, then show that the fact that the new equation is divisible by 6 follows directly from the fact that the origional equation is divisible by 6. Maybe you might get something likeXyius said:I need to prove by mathematical induction that 6 divides (k^2+5)k for all k>=1. It works for 1 but I am having trouble obtaining the k+1'st term from the k'th term. Can anyone help? Thanks!
When we say that 6 divides (n^2+5)n, it means that 6 is a factor of (n^2+5)n, which means that (n^2+5)n is divisible by 6 without any remainder.
(n^2+5)n is the number that we are trying to determine if it is divisible by 6. It is a product of two terms, n^2+5 and n, and we are checking if this product is divisible by 6.
To prove that 6 divides (n^2+5)n, we need to show that (n^2+5)n can be written as 6 multiplied by some other integer. This can be done by factoring (n^2+5)n and showing that one of the factors is 6.
No, (n^2+5)n is not always divisible by 6. For example, if n=3, then (n^2+5)n will be equal to 42, which is divisible by 6. But if n=2, then (n^2+5)n will be equal to 18, which is not divisible by 6.
One example is n=4. When n=4, (n^2+5)n is equal to 108, which is divisible by 6 since it can be written as 6 multiplied by 18.