Prove that the sum of squares of the first n natural numbers is n/6(n+1)(2n+1)

In order to do this we must follow the standard procedure for a proof by induction which is to first check a base case:Let n = 1, then the sum can be written as 1² = 1/6(1+1)(2+1) = 1 as required.
Next, assume through this check that the assumption holds for some n = k. (Where the assumption is that the sum of squared natural numbers up to n is equal to n/6(n+1)(2n+1)).
Finally, let n = k + 1 and try to show the assumption is still valid. By showing this is the case for an arbitrary n = k we can see that it will hold for all n in the natural numbers:Would show this on the whiteboard as it is a lot of numbers to type, but it works.