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

So, if $A_r = (r + 1)^2(2r^2 + 4r + 1)$ is true, then $$A_{r+1} = (r+1)^2(2r^2 + 4r + 1) + (2r + 3)^3$$ And now I can't transform the expression above into the form $$(r + 2)^2(2(r + 1)^2 + 4(r+1) + 1)$$ I tried to open these terms and got $2r^4 + 16r^3 + 47r^2 + 60r + 28$, but it seems to be a very difficult expression.

I will be grateful for any hints.

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  • $\begingroup$ Yes, they are the same expressions, but I can't see information which can help me there. $\endgroup$ Jul 20, 2018 at 11:22
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    $\begingroup$ You've expanded $ (r+1)^2(2r^2 + 4r + 1) + (2r + 3)^3$, now expand $(r + 2)^2(2(r + 1)^2 + 4(r+1) + 1)$ and check the terms $\endgroup$ Jul 20, 2018 at 11:36
  • $\begingroup$ @CalvinKhor, Ok. Thank you! I supposed that there is an efficient way to get $(r+2)^2(2(r+1)^2+4(r+1)+1)$ from $(r+1)^2(2r^2 + 4r + 1) + (2r + 3)^3$, how I did it with the sum of squares $\endgroup$ Jul 20, 2018 at 11:39
  • $\begingroup$ @E.Shcherbo possibly, yeah. I don't claim that this the most elegant way forward but it definitely gets the result $\endgroup$ Jul 20, 2018 at 11:40
  • $\begingroup$ @José Carlos Santos This question is different: it asks about the inductive step, not the base case. $\endgroup$
    – user88319
    Jul 20, 2018 at 14:17

4 Answers 4

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Expanding out quartics can be hard work, although it's a little easier if we set $r=k-1$ first. Since $2r^2+4r+1=2(r+1)^2-1$ and $A_{k-1}=k^2(2k^2-1)$, $A_{k-1}+(2k+1)^3=2k^4+8k^3+11k^2+6k+1$ while $A_k=(k^2+2k+1)(2k^2+4k+1)=2k^4+8k^3+11k^2+6k+1$. Or if even that proof that $(r+1)^2(2r^2+4r+1)+(2r+3)^3=(r+2)^2(2(r+1)^2+4(r+1)+1)$ is too much expansion to follow, you can prove these quartics are equal by checking they agree at $5$ values, e.g. $0,\,\pm 1,\,\pm 2$. This is because their difference is of degree $\le 4$, so cannot be $0$ in $5$ places unless it's constant.

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  • $\begingroup$ Are you sure that we can do it in this way? I have been thinking about your answer for a long time. Induction supposes that we get (n+1) based on n. You try to compare n-1 with n, but you didn't prove n proposition. I'm sorry if I am wrong, I'm just trying to understand it better. $\endgroup$ Jul 20, 2018 at 12:46
  • $\begingroup$ I'm proving $k$ based on $k-1$. $\endgroup$
    – J.G.
    Jul 20, 2018 at 14:22
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$$1^3 + 3^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2 + 4n + 1)$$

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

We need to show $$2(n+2)^4- (n+2)^2 - 2(n+1)^4+ (n+1)^2 =(2n+3)^3$$

Note that

$$2(n+2)^4- (n+2)^2 - 2(n+1)^4+ (n+1)^2\\= 2((n+2)^4 -(n+1)^4) -((n+2)^2 -(n+1)^2)$$

$$= 2((n+2)^2 +(n+1)^2)(2n+3) -(2n+3) \\= (2n+3)(2((n+2)^2 +(n+1)^2)-1) = (2n+3)^3$$

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The key to nearly all proofs by induction of expressions similar to this is a 6-step process:

  1. Define the proposition for $A_r$.
  2. Prove the base case. This is usually $A_1$.
  3. Consider the left-hand side of $A_r$ with the inclusion of the additional $(r+1)$-th term.
  4. Expand out all brackets
  5. Factorise again. Leverage the factor theorem to find factors one-by-one.
  6. Remember that you actually know the requisite factors, because you know that the end goal is the right hand side of $A_{r+1}$.

Thus, in your case, as it goes like this:

To Prove:

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

Proof:

Let $A_r$ be the proposition that $$1^3 + 3^3 + ... + (2r + 1)^3 = (r+1)^2(2r^2 + 4r + 1) \quad \textrm{...(Eqn 1)}$$

Clearly $A_1$ is true, as lhs = $1^3+3^3 = 28$, and rhs = $(1+1)^2(2+4+1)=28 = $lhs.

Now consider the expression

\begin{align} & 1^3 + 3^3 + ... + (2r + 1)^3 +(2r+3)^3 \\ & = \left( 1^3 + 3^3 + ... + (2r + 1)^3 \right) +(2r+3)^3 \\ & = (r+1)^2(2r^2 + 4r + 1) + (2r+3)^3 \quad \textrm{(using Eqn 1.)} \end{align}

Expanding the parentheses, we get: \begin{align} & (r^2+2r+1)(2r^2 + 4r + 1) + (8r^3+36r^2+54r+27) \\ & = (r^2+2r+1)(2r^2 + 4r + 1) + (8r^3+36r^2+54r+27) \\ & = (2r^4+8r^3+11r^2+6r+1) + (8r^3+36r^2+54r+27) $$\\ & = 2r^4+16r^3+47r^2+60r+28 \quad \quad \textrm{...(Eqn 2)} \end{align}

Now let $f(r) = 2r^4+16r^3+47r^2+60r+28 $

Since $f(-2) = 32 - 128 +188-120+28 =0$,

the factor theorem implies that $(r+2)$ is a factor of $f(r)$.

Therefore, $f(r) = (r+2) (2r^3+12r^2+23r+14)$.

Similarly, we note that if $p(r) = (2r^3+12r^2+23r+14)$, then $p(-2) = 0$.

Thus, $(2r^3+12r^2+23r+14) = (r+2) (2r^2+8r+7)$.

Therefore considering Eqn 2 again, \begin{align} & 2r^4+16r^3+47r^2+60r+28 = (r+2)^2(2r^2+8r+7) \\ & = (r+2)^2(2r^2+4r+2 + 4r+4 +1 ) \\ & = (r+2)^2(2r^2+4r+2 + 4r+4 +1 ) \\ & = \left( \overline{r+1}+1 \right)^2 \left( 2\overline{r+1}^2 + 4 \overline{r+1} +1 \right) \\ & = \textrm{rhs} \end{align}

Note This expression for the sum of odd cubes is slightly unconventional, is this refers to the sum of the first $(n+1)$ cubes. This does not change the induction proof, but readers should be careful when comparing this formula to others that they may find in math books and on the internet, which usually simply gives the formula for the first $n$ odd cubes.

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  • $\begingroup$ Thank you very much for the detailed answer. $\endgroup$ Jul 20, 2018 at 12:48
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It can be also solved using the following steps:

\begin{align} & 2r^4 + 16r^3 + 47r^2 + 60r + 28\\ & = 2r^4 + 8r^3 + 8r^3 + 8r^2 + 32r^2 + 7r^2 + 32r + 28r + 28 \\ & = (2r^4 + 8r^3 + 8r^2) + (8r^3 + 32r^2 + 32r) + (7r^2 + 28r + 28) \\ & = 2r^2(r^2 + 4r + 4) + 8r(r^2 + 4r + 4) + 7(r^2 + 4r + 4) \\ & = 2r^2(r+2)^2 + 8r(r+2)^2 + 7(r+2)^2 \\ & = (r+2)^2(2r^2 + 8r + 7) \\ & = (r+2)^2(2(r+1)^2 + 4(r+1) + 1) \\ \end{align}

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