Partitioning into groups

We note that the last term of $n^\text{th}$ group is a triangular number $T_n = \dfrac{n(n+1)}2 = l$ . The first term of $n^\text{th}$ group is thus $T_{n-1} + 1 = \dfrac{n(n-1)}2 + 1 = a$ . The number of terms in $n^\text{th}$ group is $l-a+1$ . Since the $n^\text{th}$ group is an arithmetic progression , its sum is given by:

$\begin{aligned} S & = \frac{(l-a+1)(a+l)}2 \\ & = \frac{\left(\frac{n(n+1)}2 - \frac{n(n-1)}2-1 +1\right) \left( \frac{n(n+1)}2 + \frac{n(n-1)}2+1\right)}2 \\ & = \frac{n\left(n^2+1\right)}2 \\ & = \boxed{\dfrac{n^3+n}2} \end{aligned}$

The first term of an $n^{th}$ group can be represented as $a=\dfrac{n(n+1)}{2} - (n-1)$ .

Now the common difference is always $d=1$ .

and since we know $a \space and \space d$ .

Sum of an A.P. can be given as $S=\dfrac{n}{2}[2a+(n-1)d]$ $=\dfrac{n}{2}[2(\dfrac{n(n+1)}{2} - (n-1))+(n-1) \times 1]$ $=\color{magenta}{\boxed{\dfrac{n^{3}+n}{2}}}$