sum of the first $n^2$ natural numbers closed form

Question

Before I get downvoted I am still a beginner so please bare with me. I know the summation of the first n are $\frac{n(n+1)}{2}$. Does that imply the sum of the first $n^2$ is $\frac{n^2(n^2+1)}{2}$?

Answer 1

As you know, we have that $$\sum_{k=1}^m k = 1 + 2 + \cdots + m = \frac{m(m+1)}{2}.$$This is true for any counting number (natural number) $m$. Therefore, by using this formula with $m = n^2$ for some $n$, gives $$\sum_{k=1}^{n^2} k = 1 + 2 + \cdots + n^2 = \frac{n^2(n^2+1)}{2}.$$