I was asked to give the explicit formula of: $k_n = k_{n-1} + n$ with $k_1 = 4$. I came up with the following formula: $$k_n = 3 + \sum\limits_{k=1}^n k$$ I'm wondering if adding the constant to the Sigma notation is actually legal since I couldn't find this anywhere.
EDIT: I am so sorry, the original sequence was $k_n = k_{n-1} + n$, not $k_n = k_{n-1} + 1$.
On
You need not worry. $\sum_{k=1}^nk$ is just a number, and it is perfectly legal to add $3$ and that number. "$3+$" is not a part of the "sigma notation" at all.
Having said that, your formula for $k_n$ is not right. Did you know that $\sum_{k=1}^nk=\frac{n(n+1)}{2}$? So, what you've written is $k_n=3+\frac{n(n+1)}{2}$.
On the other hand, it is pretty obvious that it is just $k_n=3+n$. Try to prove it.
Edit: After you fixed your question, it seems that your solution (to the edited question) is fine.
Writing $3 + \sum_{k=1}^n k$ is fine. Although in fact, there is a closed formula for this expression.
Just so you know, one common mistake is to write it as $$\sum_{k=1}^n k + 3.$$ I would generally avoid this because it's unclear whether it means $$\sum_{k=1}^n (k+3)$$ or $$(\sum_{k=1}^n k)+3,$$ and the two are different.