Prove by induction that $n^3 = (n^2-n+1)+(n^2-n+3)+...+(n^2+n-3)+(n^2+n-1)$

Question

I have an exercise that asks me to prove that $n^3 = (n^2-n+1)+(n^2-n+3)+...+(n^2+n-3)+(n^2+n-1)$ by induction, but I got stuck: I don't know what I can do. Could you please give me some hints? Examples: $$1^3=1 \,, 2^3 = 3+5 \,, 3^3=7+9+11 \,, 4^3=13+15+17+19$$ Note: As the exercise asks me to prove the identity $1^3 + ... + n^3 = (1+2+...+n)^2$ afterwards, I kindly ask you not to use that.

Note2: I'm aware of this identity: $ 1+3+...+2k-1 = k^2$ Thanks a lot.

Answer 1

Hint: To get from the $n$-term sum $$ (n^2-n+1) + \dots + (n^2+n-1)$$ to the $n+1$-term sum $$ ((n+1)^2-(n+1)+1) + \dots + ((n+1)^2+(n+1)-3) + ((n+1)^2+(n+1)-1)$$ we can add $(n+1)^2-n^2-1$ to each of the first $n$ terms, and then add in the last $((n+1)^2+(n+1)-1)$ term.

Answer 2

\begin{eqnarray*} \sum_{k=1}^{n} (n^2-n+2k-1)= ? \end{eqnarray*} Now use your second note.