Prove that sum of the $k$ numbers in the $k$th group = ${\frac{1}{2}\left(k(k^2+1)\right)}$. Consider an arrangement of the positive integers, grouped as shown, so that the $k$th group has $k$ elements: $(1),(2,3),(4,5,6),(7,8,9,10), \ldots$.
The expression for the sum of the $k$ numbers in the $k$th group turns out to be ${\frac{1}{2}\left(k(k^2+1)\right)}$.
However, how would you prove this? I am assuming that you would have to proof by induction, but I can't seem to construct it as of now.
 A: Let $a_{i,j}$ be the $j$'th term of the $i$'th group. The first element of the $k$'th group is the sum of the previous $1$ to $k - 1$ groups, plus $1$. Since each $i$'th group has $i$ elements, from Arithmetic progression, the first term of the $k$'th group is
$$a_{k,1} = 1 + \frac{(k-1)(1 + (k - 1))}{2} = 1 + \frac{k(k-1)}{2} \tag{1}\label{eq1}$$
Since the last term of the $k$'th group would be $1$ less than the first term of the $(k+1)$'th group, \eqref{eq1} gives it would be $\frac{k(k+1)}{2}$. Thus, the sum of the $k$'th group is
$$\begin{equation}\begin{aligned}
S_k & = \frac{k(a_{k,1} + a_{k,k})}{2} \\
& = \frac{1}{2}k\left(1 + \frac{k(k-1)}{2} + \frac{k(k+1)}{2}\right) \\
& = \frac{1}{4}k\left(2 + (k^2 - k) + (k^2 + k)\right) \\
& = \frac{1}{4}k\left(2 + 2k^2\right) \\
& = \frac{1}{2}(k(k^2 + 1))
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
A: I noticed two patterns:


*

*kth group has k numbers  

*last number of kth group is a triangular number = $\large {k(k+1) \over 2}$
First number of kth group = $\large {k(k+1) \over 2} \normalsize - k + 1$ 
Mean of kth group = $\large{k(k+1) - k + 1 \over 2} = {k^2+1 \over 2}$
Sum of kth group = $\large {k(k^2+1) \over 2}$
