prove sum of I to power I to power of two

[trim grove]

I don't know how is it proven

[grim mist]

I think it's actually:

$\ \sum_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

[mild kestrel]

adonhs
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[grim mist]

Which I believe can be proven with Induction

[trim grove]

Yeah that's it

[trim grove]

How's that like?

[grim mist]

Induction has basicslly 3 steps:

Let's say the sum is true for $n\in \mathbb{N} \cup {0}\$

First prove the statement to be true for the smallest index/number, in this case n = 0.

Then you can suppose that for some number $n\in \mathbb{N} \cup {0}\$ worked.

At last you now do this:

n -> n+1 meaning you try to prove it now for the following indices too using your assumption that it worked for some n too.