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Let $n$ be a positive integer. Prove that \[n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\]by counting the number of ordered triples $(a,b,c)$ of positive integers, where $1 \le a,$ $b,$ $c \le n,$ in two different ways.

 Nov 1, 2020
 #1
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Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

 

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

 

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

 

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

 Nov 2, 2020
 #2
avatar+111546 
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1 is a positive integer and if n=1 then the expression makes no sense.

So I suppose that is an instant disproof.

 

If it is true at all, n would have to be a positive integer greater or equal to 3

 

I have absolutely no idea what the last sentence in the question is referring to.

 Nov 2, 2020
edited by Melody  Nov 2, 2020
 #3
avatar+111546 
+2

\(n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\)

 

Prove by induction for n>=3

 

Step 1

Prove true for n=3

LHS =27

RHS=27

So the statement is true of n=3

 

Step 2

Assume true for n =k where k is some integer greater or equal to 3

So this statement is true:

 

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\ k^3=3k^2-2k+\frac{k!}{(k-3)!}\\ k^3-3k^2+2k=\frac{k!}{(k-3)!}\\ \)

 

NOW prove that it will also be true for n=k+1

i.e.   Prove

 

\((k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\ LHS=x^3+3k^2+3k+1\\~\\ RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\ RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\ RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\ sub\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\ RHS=3k^2+4k+1+(k+1)*k(k-1)\\ RHS=3k^2+4k+1+k^3-k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\)

 

So if it is true for n=k, it will be true for n=k+1

 

Step 3

The statement is true for n=3 so it is true for n=4, n=5, .....

Hence the statement is true for all integer values of n greater or equal to 3.

 

 

 

 

 

LaTex coding:

k^3=k+3k(k-1)+6\binom{k}{3}\\
k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\
k^3=3k^2-2k+\frac{k!}{(k-3)!}\\
k^3-3k^2+2k=\frac{k!}{(k-3)!}\\

 

(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\
LHS=x^3+3k^2+3k+1\\~\\
RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\
RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\
RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\
sub\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\
RHS=3k^2+4k+1+(k+1)*k(k-1)\\
RHS=3k^2+4k+1+k^3-k\\
RHS=k^3+3k^2+3k+1\\
RHS=LHS

 Nov 2, 2020
edited by Melody  Nov 2, 2020

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