Proof using induction Σ k+3 = n^2 / 2 + (7/2)n

Question

So I have the following problem

prove: Σ(n)(k+1) k+3 = n^2 / 2 + (7/2)n 
P(1) = 1 + 3 = 1^2 / 2 + 7 /2
P(1) = 4 = 4
So I assume it's true for n and attempt to prove it's true for n+1:
(n+1) + 3 = (n+1)^2 / 2 + (7/2)(n+1)
n+4 = (n^2 + 2n + 1) /2 + (7/2)(n+1)

and now I'm stuck. I have no idea how to make the functions equal. I can't really see a mistake either.

Answer 1

$\sum\limits_{k=1}^n (k + 3) = \frac{n^2 + 7n}{2}$
For the base case, $n=1$. $$1+3=\frac{8}{2}=4$$
Let us assume that it is true for some $n=p$, then $$\sum\limits_{k=1}^p (k + 3) = \frac{p^2 + 7p}{2}$$ Now it has to be true for all $n=p+1$, then $$\sum\limits_{k=1}^{p+1} (k + 3) = \frac{(p+1)^2 + 7(p+1)}{2}$$ $$\sum_{k=1}^p(k+3)+p+4=\frac{(p+1)^2+7p+7}{2}$$ $$p+4=\frac{(p+1)^2+7p+7}{2}-\frac{p^2+7p}{2}$$ $$p+4=\frac{(2p+8}{2}$$ $$p+4=p+4$$ Which is true. Hence proved.