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I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$

I already did the induction steps:

Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)

Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$

To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$

My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.

Simplifying the left and right part of the statement does not help:

Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $

Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$

So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$

That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$

Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$

Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$

Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$

Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?

Martin Sleziak
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Laurence
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6 Answers6

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Show that $S_n-S_{n-1}=n^3$ (or if you prefer, $S_n=S_{n-1}+n^3$).

Indeed, $$\left(\frac{n(n+1)}2\right)^2-\left(\frac{(n-1)n}2\right)^2=n^2\left(\frac{n+1}2+\frac{n-1}2\right)\left(\frac{n+1}2-\frac{n-1}2\right).$$

Note that induction is unnecessary. As both members are quartic polynomials, it suffices to prove equality for five distinct values of $n$:

$$0=0,\\1=1,\\9=3^2,\\36=6^2,\\100=10^2.$$

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Hint

It's easier to begin by the end

$$\left(\frac{(n+1)(n+2)}{2}\right)^2-\frac{n(n+1)}{2}^2=$$

$$\left(\frac{n+1}{2}\right)^2\left((n+2)^2-n^2\right)=$$

$$\left(\frac{n+1}{2}\right)^2(4(n+1))=$$

$$(n+1)^3.$$ qed.

hamam_Abdallah
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Take your "to be proven" equation and divide through by $(k+1)^2$ and multiply through by $4$. This gives $k^2+4(k+1)=(k+2)^2$ as the equation to be proven, and this is trivial.

diracula
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\begin{align} \underbrace{1^3+2^3+\ldots+n^3}_{\left[\frac{n(n+1)}{2} \right]^2}+(n+1)^3 =& \left[\frac{n(n+1)}{2} \right]^2+(n+1)^3 \\ =& \frac{n^2\color{red}{(n+1)^2}}{4} + (n+1)\color{red}{(n+1)^2} \\ =& \left\lgroup \frac{n^2}{4} + (n+1) \right\rgroup \color{red}{(n+1)^2} \\ =& \left\lgroup \frac{n^2 +4(n+1)}{4} \right\rgroup \color{red}{(n+1)^2}\\ =&\left\lgroup \frac{(n+2)^2}{4} \right\rgroup \color{red}{(n+1)^2} \\ =&\left\lgroup \frac{(n+2)}{2} \right\rgroup^2 \color{red}{(n+1)^2} \\ =&\left\lgroup \frac{\big(\color{red}{(n+1)}+1\big)}{2} \right\rgroup^2 \color{red}{(n+1)^2} \\ =&\left[ \frac{\color{red}{(n+1)}\big(\color{red}{(n+1)}+1\big)}{2} \right]^2 \end{align}

Elias Costa
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After your $" to be proven "$ step, follow the rules below: 1. Take $ (k+1)^{2} $ common so we have then $ (k+1)+\frac {k}{4} ^{2} $. 2. Then simplify the result to get $ \frac { k^{2}+4k+4}{4} $. 3.This is equivalent to $\frac {(k+2)}{2}^{2} $ 4. Proof finished as we have our result equal to $ \frac {(k+1)(k+2)}{2} ^{2} $.

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First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}k^3=(1(1+1)/2)^2$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}k^3=(n(n+1)/2)^2$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}k^3=$

$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$

$\color\red{(n(n+1)/2)^2}+(n+1)^3=$

$((n+1)(n+2)/2)^2$

Please note that the assumption is used only in the part marked red.

barak manos
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