A.P.: Sum of numbers in a particular group Question: A series of odd positive integers are divided into the following groups $(1), (3,5), (7,9,11),\dotsm$ . Prove that the sum of numbers in the $n^{\text{th}}$ group is $n^3$.
My attempt:
It is observed that the number of terms in the $n^{\text{th}}$ group is equal to $n$. The common difference is same in all groups and is equal to 2. If we do not consider the grouping, then the resulting sequence $1,3,5,\dotsm$ is in A.P. with first term $t_1$ equal to 1 and common difference equal to 2. Now the first term of the first group is equal to $t_1=t_{1+0}$, that of the second group is equal to $t_2=t_{1+1}$, that of the third group is equal to $t_4=t_{1+1+2}$, that of the fourth group is equal to $t_7=t_{1+1+2+3}$ and so on. Thus the first term of the $n^{\text{th}}$ group is equal to $t_{1+S_{n-1}}=t_{1+\frac{(n-1)n}{2}}=t_{\frac{n^2-n+2}{2}}=t_1+(\frac{n^2-n+2}{2}-1)2=t_1+\frac{n^2-n}{2}\times2=t_1+n^2-n=1+n^2-n$
Here $S_{n-1}$ denotes the sum of first $(n-1)$ natural numbers.
Therefore the first term of the $n^{\text{th}}$ group is equal to $n^2-n+1$.
Sum of all numbers in the $n^{\text{th}}$ group $=S^{'}_n=\frac{n}{2}[2(n^2-n+1)+(n-1)2]=\frac{n}{2}[2n^2-2n+2+2n-2]=n^3$.
My problem: I am looking for other methods to prove this result.

To form $n$ groups you need $1+2+3+\cdots+n=\frac{n(n+1)}{2}$ terms. To form $n-1$ groups you need $1+2+3+\cdots+n-1=\frac{n(n-1)}{2}$ terms. To get the sum of the terms in the $n$-th group you sum all the terms from the $1$st group to the $n$th and subtract the sum from the $1$st group to the $n-1$th which is
$$S_{n(n+1)/2}-S_{n(n-1)/2}=\frac{n(n+1)}{4}\left(2+2\left(\frac{n(n+1)}{2}-1\right)\right)-\frac{n(n-1)}{4}\left(2+2\left(\frac{n(n-1)}{2}-1\right)\right)=\frac{n(n+1)}{2}(\frac{n(n+1)}{2})-\frac{n(n-1)}{2}(\frac{n(n-1)}{2})=\frac{n^2}{4}((n+1)^2-(n-1)^2)=n^3$$
