A rhombus is constructed out of natural numbers with the length of the side being 1,2,3,... in a way specified below. Find the sum of the n'th rombs numbers and prove the correctness of the result with the way of mathematical induction.
So there is no problem with finding the n'th rombs sum. It's n^3. But I'm having trouble with writing the left side of the equasion to use the mathematical induction.
.... = n^3
.... = (n+1)^3
To use induction, note each "rhombus" is constructed from the previous one by adding a diagonal on the lower left & lower right sides, with the very bottom number repeated between them. In particular, going from $k$ to $k+1$, the diagonals being added go from $k + 1$ to $2k + 1$, with the $2k + 1$ only occurring once. For example, from the third to the fourth rhombii, you have the diagonals of $4$ to $7$ added to the previous rhombus.
To use this in your inductive step, first let
$$S(n) = n^3 \tag{1}\label{eq1A}$$
You can easily confirm this works for $n = 1$ since you have just $1$ and that's equal to $1^3$. Assume \eqref{eq1A} gives the sum for all $n \le k$ for some $k \ge 1$. For $n = k + 1$, the sum (using the sum of an arithmetic series of $a_1$ to $a_n$ is $\frac{n(a_1 + a_n)}{2}$) would be
$$\begin{equation}\begin{aligned} S(k) + 2\sum_{i=k+1}^{2k}i + (2k + 1) & = k^3 + 2\left(\frac{k((k+1) + 2k)}{2}\right) + (2k + 1) \\ & = k^3 + k(3k + 1) + (2k + 1) \\ & = k^3 + (3k^2 + k) + (2k + 1) \\ & = k^3 + 3k^2 + 3k + 1 \\ & = (k + 1)^3 \\ & = S(k+1) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This shows \eqref{eq1A} also gives the sum for $n = k + 1$. Thus, by induction, \eqref{eq1A} gives the sum of the numbers of the $n$'th rhombus for all $n \ge 1$.