$$\sum_{i=0}^n(2i+3)^2=\frac{(n+1)(4n^2+20n+27)}{3}$$
Hey! :) I need help with this problem, I'm kinda silly and I've got a problem with this, I don't even know how to start. If you guys can help me that would be amazing. I'm in my first year of university and I don't know how to do it. :(
As mentioned in the comments, you can prove this either with so-called Mathematical Induction or just calculate the expression on the LHS. I will try show the second approach here. $$\sum_{i=0}^n(2i+3)^2=\sum_{i=0}^n(4i^2+12i+9)=4\underbrace{\sum_{i=0}^ni^2}_{S_1}+12\underbrace{\sum_{i=0}^ni}_{S_2}+\underbrace{\sum_{i=0}^n9}_{S_3}$$ So, now we can notice that $S_1$ is just the sum of the squares of first $n$ natural numbers and $S_2$ is an arithmetic progression. Using the known formulas (which you can find the proofs by clicking on the hyperlinks above), we obtain that $$\sum_{i=0}^n(2i+3)^2=4\sum_{i=0}^ni^2+12\sum_{i=0}^ni+\sum_{i=0}^n9=4\frac{n(n+1)(2n+1)}{6}+12\frac{n(n+1)}{2}+9(n+1)=\frac{(n+1)(4n^2+20n+27)}{3}$$ as desired.
