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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.

GGG
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    Yes this is the way to find the constants $a$ , $b$ , $c$ and then prove the resulting identity by induction . –  Jan 10 '16 at 16:25
  • @ComplexPhi But when we are using induction we verify the base case first. This time we're assuming it's true for the base case to begin with. This is what's confusing me. – GGG Jan 10 '16 at 16:28
  • So we don't start the induction because we don't know yet what we want to prove . First find the $a,b,c$ such that the identity will work and then prove it (using induction ) for those specific $a,b,c$ . Hope you understand now . –  Jan 10 '16 at 16:29
  • @ComplexPhi I think I understand now. Thank you. – GGG Jan 10 '16 at 16:36

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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 $a,b,c$ )

  • Thanks, I think I understand now. But I still don't understand the "alternatively" part. Could you explain what they mean, if you don't mind please? – GGG Jan 10 '16 at 16:40
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    @user276387 Of course . They say to start directly with the induction . The base case needs to hold so you have a condition on $a,b,c$ . Then when you do the inductive step you'll get other conditions on $a,b,c$ for the step to hold . This way you can find $a,b,c$ directly in the inductive proof while with this method you first find what $a,b,c$ should be and then prove that indeed they work using induction . –  Jan 10 '16 at 16:43
  • Many thanks again! – GGG Jan 10 '16 at 16:51