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

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

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