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$1^3 + 2^3 + \cdot \cdot \cdot+ n^3 = $ $[ \frac{n(n+1)}{2}]^{2} $

$\text{My question for this problem is that I got stuck at a certain point}$

$\text{and I do not know where to go. This is what I know to prove by PMI.}$

(i) $1 ∈ S$

(ii) $ \text{for all } n ∈ ℕ \text{ if } n∈ ℕ , \text{ then } n+1 ∈ S. \text{ Then } S = ℕ $

$\text{ A set of natural numbers is called an inductive set iff it has proberty that whenever } n∈ ℕ, \text{then } n+1 ∈ S.$

(iii) $ \text{By PMI true for every } n∈ ℕ$

$\color{maroon}{Proof :}$

$\text{(i)} \; \text{ The statement is true for } n=1 \text{ because} $ $[ \frac{1(1+1)}{2}]^{2} = 1$

$ 1= 1$

$\text{(ii)} \; $$\text{Assume that for some } n∈ ℕ $

$1^3 +2 ^3 + \cdot \cdot \cdot+ n^3 = [ \frac{n(n+1)}{2}]^2 $

$ \text{Let } n = n+1$

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

= $1^3 + 2^3 + \cdot \cdot \cdot+ n^3 + (n+1)^3 = [ \frac{n(n+1)}{2}]^2 + (n+1)^3 $

$\text{This is where I get stuck at I do not know how to proceed. }$

$\text{ I am a bit confused by the whole process of PMI}$

$\text{any advice on how I can procced would be gladly appreciated.}$

Jon
  • 1,920

5 Answers5

1

Under ii, "let $n=n+1$" is not what you want. You should use the assumption to say $$1^3 +2 ^3 + \bullet \bullet \bullet + n^3 +(n+1)^3= [ \frac{n(n+1)}{2}]^2+(n+1)^3$$ Now you want to work on the right to show it is $$[\frac{(n+1)(n+2)}2]$$ which establishes the formula you want because it fits the pattern for $n+1$. I recommend Arturo Magidin's answer here for a good explanation for how to think about induction.

Community
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Ross Millikan
  • 374,822
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$$1^{ 3 }+2^{ 3 }+{ 3 }^{ 3 }...+n^{ 3 }+{ \left( n+1 \right) }^{ 3 }={ \left[ \frac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }+{ \left( n+1 \right) }^{ 3 }=$$ $$={ \left( n+1 \right) }^{ 2 }\left[ \frac { { n }^{ 2 } }{ { 2 }^{ 2 } } +\left( n+1 \right) \right] ={ \left( n+1 \right) }^{ 2 }\frac { { n }^{ 2 }+4n+4 }{ 4 } =$$ $$=\frac { { \left( n+1 \right) }^{ 2 }{ \left( n+2 \right) }^{ 2 } }{ 4 } ={ \left[ \frac { \left( n+ \right) \left( n+2 \right) }{ 2 } \right] }^{ 2 }$$

haqnatural
  • 21,578
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We make the assumption, like you did, that for some $n$ we have $$\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2\qquad (1)$$ Then if we can show that, by making this assumption, that $$\sum_{k=1}^{n+1}k^3\qquad \text{and}\qquad \sum_{k=1}^{1}k^3$$ are both of the form $(1)$ then that means we have that for any given $n$, the value $n+1$ also satisfies that equation. Thus since we have $n=1$ works, we have $n+1=2$ works; set $n=2$ and get $n+1=3$ works and so on to infinity. So assuming that $(1)$ holds for some $n$ consider $$\sum_{k=1}^{n+1}k^3$$ Upon which we do some manipulation: $$\sum_{k=1}^{n+1}k^3=\sum_{k=1}^{n}k^3+(n+1)^3$$ Using $(1)$ which we assumed to be true for $n$: $$=\left(\frac{n(n+1)}{2}\right)^2+(n+1)^3=\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4}$$ $$=\frac{n^2(n+1)^2+4(n+1)^3}{4}$$ $$=\frac{(n+1)^2[n^2+4n+4]}{4}\qquad \text{factor...}$$ $$=\frac{(n+1)^2(n+2)^2}{4}=\left(\frac{(n+1)(n+2)}{2}\right)^2$$ which is $(1)$ evaluated at $n+1$ thus if $(1)$ holds for $n$ it holds for $n+1$. Then we just have to check that it holds for $n=1$ which is true because $$\sum_{k=1}^{1}k^3=1^3=\left(\frac{1\cdot(1+1)}{2}\right)^2$$

Will Fisher
  • 5,112
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Your first steps are correct.

now you need to show:

$1^3 + 2^3 \cdot + n^3 + (n+1)^3 = [\frac{(n+1)(n+2)}{2}]^2$

based on the inductive hypothesis:

$1^3 + 2^3 \cdot + n^3 = [\frac{(n)(n+1)}{2}]^2$

$[\frac{(n)(n+1)}{2}]^2 + (n+1)^3$

$\frac{n^4 + 2n^3 + n^2}{4} + (n+1)^3\\ \frac{n^4 + 2n^3 + n^2 + 4n^3 + 12 n^2+ 12 n + 4}{4}\\ \frac{n^4 + 6n^3 + 13n^2 + 12 n + 4}{4}\\ \frac{(n+1)(n^3 + 5n^2 + 8n + 4)}{4}\\ \frac{(n+1)^2(n^2 + 4n + 4)}{4}\\ \frac{(n+1)^2(n + 2)^2}{4}\\ \big(\frac{(n+1)(n+2)}{2}\big)^2$

Doug M
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Hint: expand everything, combine terms, and then factor again:

\begin{align*} \left[n\left(n+1\right)/2\right]^{2}+\left(n+1\right)^{3} & =n^{4}/4+n^{3}/2+n^{2}/4+n^{3}+3n^{2}+3n+1\\ & =n^{4}/4+3n^{3}/2+13n^{2}/4+3n+1\\ & =\cdots\\ & =\left(n+1\right)^{2}\left(n+2\right)^{2}/4\\ & =\left[\left(n+1\right)\left(n+2\right)/2\right]^{2} \end{align*}

parsiad
  • 25,154