<s>[INST] Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$
 [/INST] Let us find the general term of $1,3,7,13,\cdots$
As the next consecutive differences are $2,4,6,\cdots$
$$T_n=an(n-1)+b(n-1)(n-2)+cn(n-2)$$
$n=1\implies1=c(1-2),c =-1$
$n=2\implies3=2a$
$n=3,7=6a+2b+3c=3(3)+2b+3(-1),2b=1$
$$\implies T_n=\cdots=n^2-n+1$$
We need  the sum of $n$ terms from $k=n^2-n+1$ to $(n+1)^2-(n+1)+1-1$ with the common difference $=2$
which is $$n(n^2-n+1)+\sum_{k=1}^n(2k)=n^3-n^2+n+\dfrac n2(2+(n-1)2)=?$$
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