Proof by induction that $17n^3+103n$ is a multiple of 6 Problem use induction to prove that $17n^3+103n$ is a multiple of 6.
Im new to learning proof by induction and was wondering if my proof would be acceptable for my maths test coming up.
Proof. base case $n=1 = 120$ $17k^3+103k=6a$ for some $a$. then we prove $k+1$ $$17(k+1)^3+103(k+1)=6b$$ for some b. we expand and get $17k^3+51k^2+51k+17+103k+103=6b$ combine like terms and substitute $17k^3+103$ for $6a$ then we get $$6a+51k^2+51k+120=6b$$ simplify to $51n(n+1)$ then we see that $n(n+1)$ has to multiply to become an even number so 51n(n+1) is also a multiple of six QED.
would this proof work or are there any better methods.

Your base case is correct: $17(1)^3+103(1)=120$ which is divisible by $6$.
For the inductive step we assume $17k^3+103k$ is divisible by $6$ and consider $17(k+1)^3+103(k+1)$.
Expanding this we get $$17(k+1)^3+103(k+1)=17(k^3+3k^2+3k+1)+103k +103=(17k^3+103k)+51k^2+51k+120$$
We know that $17k^3+103k$ is divisible by $6$ because of the induction hypothesis. We also know that $120$ is divisible by $6$ so what we really have to show is that $51k^2+51k$ is divisible by $6$.
$51k^2+51k=51(k^2+k)$ and $51$ is divisible by $3$ so we only need to show that $k^2+k$ is even.
Now, $k$ and $k^2$ are either both even or both odd. Since the sum of two even numbers and the sum of two odd numbers are even, $k^2+k$ is even so $51k^2+51k$ is divisible by $6$.
Therefore, by induction, $17n^3+103n$ is divisible by $6$ for all $n$.
