2

I'm trying to use mathematical induction to prove that $n^3+5n$ is divisible by $6$ for all $n \in \mathbb{Z}^+$.

I can only seem to show that it is divisible by $3$, and not by $6$. This is what I have done:

Let $f(n) = n^3+5n$.

Basis Step: When $n=1$, $f(n)= 6$ and clearly $6$ divides $6$.

Assumption Step: Assume that $6$ divides $f(n)$ when $n=k$.

Inductive Step: Consider $f(k+1)-f(k)$: $$f(k+1)-f(k) = [(k+1)^3+5(k+1)]-[k^3+5k]$$ $$=3k^2+3k+6$$ $$=3(k^2+k+2)$$ It follows that $f(k+1) = f(k)+3(k^2+k+2)$. I really wanted a common factor of $6$.

Anon Anon
  • 23

2 Answers2

4

$$3(k^2+k+2)=6\cdot\frac{k(k+1)}2+6$$

Now $k(k+1)$ being the product of two consecutive integers is even for integer $k$

Non-inductive method:

$$n^3+5n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers }}+6n$$

See The product of n consecutive integers is divisible by n factorial

Community
  • 1
lab bhattacharjee
  • 274,582
0

From your proof... Note that if k is odd, k^2 + k + 2 is even and hence divisible by 2, and that the same is true if k is even. Therefore k^2 + k + 2 is divisible by 2 for all k. this gives you the extra factor of 2 you need to get a factor of 6.

Brian
  • 1