Find constants $a$, $b$ and $c$ such that for all $n \in \mathbb{N}$ $~~\sum_{k=1}^{n}k(k-\frac{1}{3}) = \frac{n}{6}(an^2+bn+c)$
Hints: you may want to find $a, b$ and $c$ from the condition that the statement is true for $n = 1, 2, 3$. You will then need to prove by induction that it holds for all $n ∈ \mathbb{N}$. Alternatively, write down the proof by induction for general $a, b$ and $c$ and obtain the required conditions on $a, b$ and $c$ from the fact that the basis step in the proof by induction is true and the induction step must be valid for all $n \in \mathbb{N}$.
My concern is how is my proof valid if right from the start I assume it's true for $n=1, 2, 3$ to find the values of $a, b, c$? Could someone please explain this.
Here's the explanation :
We want the identity :
$$\sum_{k=1}^{n} k(k-\frac{1}{3} )=\frac{n}{6} (an^2+bn+c)$$ to be true for every $n$ .
In particular it needs to be true also for $n=1,2,3$ .
This way we can find what the constants need to be (by solving the system of three equations )
After we found what $a,b,c$ should be we can start the induction on $n$ and prove that indeed it works for every $n$ (for those particular constants $a,b,c$ )