I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.
I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.
Hello jess0517
We have to prove that $\displaystyle \displaystyle \tfrac13n^3+\tfrac12n^2+\tfrac16n$ is an integer for all $\displaystyle \displaystyle n \in \mathbb{N}$.
First we factorise:
$\displaystyle \displaystyle \tfrac13n^3+\tfrac12n^2+\tfrac16n=\tfrac16n(2n^2+3 n+1)$So we must prove that $\displaystyle \displaystyle n(n+1)(2n+1)$ is a multiple of 6; in other words, that it is a multiple of 2 and of 3.$\displaystyle \displaystyle =\tfrac16n(n+1)(2n+1)$
First, note that one of $\displaystyle \displaystyle n$ and $\displaystyle \displaystyle (n+1)$ is even. So $\displaystyle \displaystyle n(n+1)(2n+1)$ is even.
Next, note that $\displaystyle \displaystyle n$ is either a multiple of 3, in which case there is nothing left to prove; or it leaves a remainder 1 or 2 when divided by 3.
Case I: $\displaystyle \displaystyle n$ leaves a remainder 1; in which case $\displaystyle \displaystyle n = 3m + 1, m \in \mathbb{N}$. Therefore
$\displaystyle \displaystyle 2n+1 = 6m + 3$and is therefore a multiple of 3.
Case II: $\displaystyle \displaystyle n$ leaves a remainder 2; in which case $\displaystyle \displaystyle n = 3m +2, m \in \mathbb{N}$. Therefore
$\displaystyle \displaystyle n+1 = 3m + 3$and is therefore a multiple of 3.
This concludes the proof.
Grandad
P(k)
$\displaystyle \frac{1}{3}k^3+\frac{1}{2}k^2}+\frac{1}{6}k$ is an integer ?
P(k+1)
$\displaystyle \frac{1}{3}(k+1)^3+\frac{1}{2}(k+1)^2+\frac{1}{6}( k+1)$ is also an integer ?
Try to prove that P(k) being true causes P(k+1) to be true.
Proof
$\displaystyle \frac{1}{3}(k+1)^3+\frac{1}{2}(k+1)^2+\frac{1}{6}( k+1)=\frac{1}{3}\left(k^3+3k^2+3k+1\right)+\frac{1 }{2}\left(k^2+2k+1\right)+\frac{1}{6}\left(k+1\rig ht)$
$\displaystyle =\left(\frac{1}{3}k^3+\frac{1}{2}k^2+\frac{1}{6}k\ right)+\left(k^2+k+\frac{1}{3}+k+\frac{1}{2}+\frac {1}{6}\right)$
$\displaystyle =\left(\frac{1}{3}k^3+\frac{1}{2}k^2+\frac{1}{6}k\ right)+\left(k^2+2k+\frac{2}{6}+\frac{3}{6}+\frac{ 1}{6}\right)$
$\displaystyle =\left(\frac{1}{3}k^3+\frac{1}{2}k^2+\frac{1}{6}k\ right)+\left(k^2+2k+1\right)$
$\displaystyle \left(\frac{1}{3}k^3+\frac{1}{2}k^2+\frac{1}{6}k\r ight)+\left(k+1)^2$
If k is an integer, so is k+1 and (k+1)(k+1).
Hence, if P(k) is true, it causes P(k+1) to be true.
All that remains is to test the proposition P(k) for the first integer.