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

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 $$\frac{n^2(n^2+1)}{2}$$?

• In a word, yes. – Angina Seng Apr 10 '19 at 6:48
• Why do you say yes? I thought that the closed-form expression for the sum of the first $n^2$ numbers was $$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ No? – Michael Rybkin Apr 10 '19 at 6:50
• @MichaelRybkin No. That's the sum of the first $n$ squares. The OP is asking for the sum of the first $n^2$ numbers. – José Carlos Santos Apr 10 '19 at 6:54
• Oh, I see what you mean now. – Michael Rybkin Apr 10 '19 at 7:12

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 $$\sum_{k=1}^{n^2} k = 1 + 2 + \cdots + n^2 = \frac{n^2(n^2+1)}{2}.$$