Proving a simple algebra equation

Question

I have to prove the equality: $$1^3 + 2^3 + · · · + (n − 2)^3 + 2^3 + n^3 = (1 + 2 + · · · + (n − 2) + 2 + n)^2$$

I already know that: $$1^3+2^3+3^3...+n^3=(1 + 2 + 3...+n)^2$$

How can I derive the second from the first? Perhaps not all through algebraic manipulation, but some logic, because the algebra seems tedious in this equation.

Answer 1

Let us call the left-hand side $L$ and the right-hand side $R$. Then

$$L=1+2^3+\ldots+(n-1)^3+n^3+\left[2^3-(n-1)^3\right]=\left[\dfrac{n(n+1)}{2} \right]^2+\left[2^3-(n-1)^3 \right]$$

The right-hand side is given by: $$R = \left[1+2+...+(n-2)+(n-1)+n+(2-(n-1))\right]^2=\left[\dfrac{n(n+1)}{2} +(2-(n-1))\right]^2$$

$$=\left[\dfrac{n(n+1)}{2} \right]^2+2\dfrac{n(n+1)}{2} (2-(n-1))+(2-(n-1))^2$$

Now, try to calculate $L-R$ if it is $0$ then you are done. But as many users pointed out the identity is not correct.

Answer 2

Standard induction works. Easy to verify for $n=2,3$. We will work with first difference. The form of induction we'll use is:

Induction: Given two sequences $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$, if $a_1=b_1$ and, for all $n≥1$ $a_{n+1}-a_n=b_{n+1}-b_n$ then $a_n=b_n$ for all $n\in \mathbb N$.

Given your expression, let the left hand be $L_n$, the right hand $R_n$.

We compute:

$$L_{n+1}-L_n=(n-1)^3+(n+1)^3-n^3=n^3+6n$$

$$R_{n+1}-R_n=(1+2+\cdots+ (n-2)+(n-1)+2+(n+1))^2-(1+2+\cdots +(n-2)+2+n)^2$$

We remark that $$1+\cdots + k=\frac {k(k+1)}2$$ whence $$R_{n+1}-R_n=\big(\frac {n(n+1)}2+3\big)^2-\big(\frac {n(n-1)}2+3\big)^2=n^3+6n$$

and we are done.

Answer 3

Let $A = 1 + 2 + \cdots + (n-1)$. Then $A = (n-2)(n-1)/2$ and the identity reads

$$A^2 + 2^3+n^3 = (A+2+n)^2.$$

An $A^2$ cancels and we have

$$2^3+n^3 = 2A(2+n) + (2+n)^2$$

or

$$(2+n)(2^2-2n+n^2) = (n-1)(n-2)(2+n)+(2+n)^2.$$

Cancel the factor of $n+2$.

$$2^2-2n+n^2 = (n-1)(n-2)+(n+2) = n^2 - 3n+ 2 +2+n = n^2-2n +4.$$

Reverse the step and you have a nice proof.