0

Using this formula:

$1^3+2^3+⋯+n^3=[\frac{n(n+1)}{2}]^2$

Prove:

$1^3+3^3+⋯+(2n+1)^3=(n+1)^2(2n^2+4n+1)$

I had a hard time trying to prove this, so I'll be glad if someone could help me.

This is a exercise from the book "What is Mathematics?" from Richard Courant

Community
  • 1

3 Answers3

1

That is pretty simple: $$1^3+3^3+\ldots+(2n+1)^3 = \sum_{k=1}^{2n+1}k^3 -\left[2^3+4^3+\ldots(2n)^3\right] \tag{1}$$ leads to: $$ 1^3+3^3+\ldots+(2n+1)^3 = \left(\frac{(2n+1)(2n+2)}{2}\right)^2-8\left(\frac{n(n+1)}{2}\right)^2 \tag{2}$$ and you just have to collect a $(n+1)^2$ from the RHS of $(2)$ and simplify.
The correct outcome is $(n+1)^2(2n^2+4n+1)$, without any extra square, also because $1^3+\ldots+(2n+1)^3$ has to be a fourth-degree polynomial in $n$.

Jack D'Aurizio
  • 353,855
  • Sorry, you're right.

    But my problem is also the simplification"! I tried to expand the monomials, but I couldn't

     –  Jun 18 '17 at 01:48
  • By dividing the RHS of $(2)$ by $(n+1)^2$ you are left with $$ (2n+1)^2-8\frac{n^2}{4} = (4n^2+4n+1)-(2n^2) = (2n^2+4n+1).$$ – Jack D'Aurizio Jun 18 '17 at 01:56
  • Wow! This was magic!

    Ok,ok,I'm slow in this, so lets make step by step.

    So you notice the two parts by having the $(n+1)^2. $

    Then you multiply by $1$, in the case $(n+1)^2 /(n+1)^2. $

    Simplify then multiply! I'm right?

     –  Jun 18 '17 at 02:07
  • @MarcosConci: I am just exploiting $(ab)^2 = a^2 b^2$ and $a^2 c - a^2 d = a^2(c-d)$, but that is clearly the same as proceeding as you suggest. – Jack D'Aurizio Jun 18 '17 at 02:50
1

Hint: \begin{align} &1^3 + 3^3 + \dotsb + (2n+1)^3 \\ &= \big(1^3 + 2^3 + 3^3 + 4^3 + \dotsb + (2n)^3+(2n+1)^3\big) - \big(2^3 + 4^3 + \dotsb + (2n)^3\big) \\ &= \big(1^3 + 2^3 + 3^3 + 4^3 + \dotsb + (2n+1)^3\big) - 2^3\big(1^3 + 2^3 + \dotsb + n^3\big) \end{align}

Chappers
  • 67,606
0

Based on @Jack D'Aurizio, I will try to answer my question.

So the problem says that:

$$1^3+2^3+3^3+\ldots+n^3 =\left[\frac{n(n+1)}{2}\right]^2\tag{1}$$

And if it was true, it will also be true for:

$$1^3+2^3+3^3+\ldots+(2n+1)^3+ =\left[\frac{n(n+1)}{2}\right]^2\tag{2}$$

But (2) also includes de even number, and the question ask for the odds ones.

$$ 1^3+3^3+\ldots+(2n+1)^3 = \left(\frac{(2n+1)(2n+2)}{2}\right)^2-8\left(\frac{n(n+1)}{2}\right)^2 \tag{4}$$

Where the even ones are express by:

$2^3+4^3+6^3+\ldots+(2n)^3+=\left[\frac{2n(2n+1)}{2}\right]^2\to 2^3(1^3+2^3+\ldots+n^3)=8\left(\frac{n(n+1)}{2}\tag{3}\right)^2$

Now we have to simplify our expression by dividing then multiplying the RHS of (4) by $(n+1)^2$ (That is multiplying by one, which don't change our expression on the both sides)

$$\frac{(n+1)^2}{(n+1)^2}\Biggl[\left(\frac{(2n+1)(2n+2)}{2}\right)^2-8\left(\frac{n(n+1)}{2}\right)^2\Biggr]$$

$$(n+1)^2\Biggl[(2n+1)^2-8\left(\frac{n^2}{4}\right)\Biggr]$$

That leads us finally to:

$$1^3+3^3+⋯+(2n+1)^3=(n+1)^2(2n^2+4n+1)$$

My thanks to those who answer me!