So there exist a lot of similar questions which ask the $n$th term of   

$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T\_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider each number as a single term. I tried making a pyramid of terms and got this

$1^2\\
1^2~2^2\\
1^2~2^2~3^2\\
1^2~2^2~3^2~4^2\\
...$

I got the $n$th term as$$T\_n= (n\mod(m(m-1)/2)^2$$
Where $m$ is the row number.

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> But now I am stuck, is there a way to eliminate $m$ and write the general term only in terms of $n$?
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