Prove by induction $\sum_{i=1}^ni^2=1^2+2^2+3^2+…+n^2=\frac{n(n+1)(2n+1)}{6}=Θ(n^3)$

Question

I am trying to prove $$\sum_{i=1}^ni^2=1^2+2^2+3^2+…+n^2=\frac{n(n+1)(2n+1)}{6}=Θ(n^3)$$

I understand when it comes to big-theta that f(n)= $$\sum_{i=1}^ni^2=1^2+2^2+3^2+…+n^2=\frac{n(n+1)(2n+1)}{6}$$ and g(n) = $$Θ(n^3)$$

I know that Θ(g(n)) = { f(n) : ∃ positive constants c1, c2 and n0, for ∀n ≥ n0 0 ≤ c1 g(n) ≤ f(n) ≤ c2 g(n)

My issue is how to solve this by induction. I've learned induction but I am unsure of how to do this with summation and big-theta

I know I have to use k+1 but how do I use it when it comes to this question?

Basis:

S1=1=(1)(1)+1(2(1)+1)/6 = 2(3)/6 =1

So basis is true. Or is this even the right basis for solving by induction?

Hypothesis :

Sk = $$\sum_{i=1}^ki^2=1^2+2^2+3^2+…+k^2=\frac{kk+1(2k+1)}{6} < c2^3$$

Induction :

Sk+1 = $$\sum_{i=1}^ki^2=1^2+2^2+3^2+…+k+1(k+1)^2=\frac{(k+1)(k+2)(2k+2)}{6} < c2^3$$

I am not sure if I did any of this correctly if not can someone show me how?

Answer 1

Your basis is correct. The inductive step would be:

$\begin{align}1^2+2^2+3^2+....+k^2+(k+1)^2&=S_k+(k+1)^2\\&=\frac {k(k+1)(2k+1)}{6} +(k+1)^2\\&=(k+1)\left(\frac {k(2k+1)}{6}+(k+1)\right)\\&=(k+1)\left(\frac {2k^2+7k+6}{6}\right)\\&=(k+1)\left(\frac {(k+2)(2(k+1)+1)}{6}\right)\\ &=S_{k+1}\end{align}$
Hence proved.