Question 1042569
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Prove Qn:  {{{1^3+2^3+""*""*""*""+n^3}}}{{{""=""}}}{{{(1+2+""*""*""*""+n)^2}}}

First we will prove a well-known formula for the sum 
of the first n positive integers:

Pn: {{{1+2+""*""*""*""+n}}}{{{""=""}}}{{{(n(n+1))/2}}}

Then we can square both sides and get:

{{{(1+2+""*""*""*""+n)^2}}}{{{""=""}}}{{{((n(n+1))/2)^2}}}

then we will prove

{{{1^3+2^3+""*""*""*""+n^3}}}{{{""=""}}}{{{( (n(n+1))/2 )^2}}}

Then what we are to prove will be immediate.

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First induction proof for 

Pn:  {{{1+2+""*""*""*""+n}}}{{{""=""}}}{{{(n(n+1))/2}}}
 
First let's see what Pk+1 would be:

[That's always the first thing to do.  Before you 
start an induction proof,  you should calculate Pk+1 
to see where you're headed]:

To do that, replace n by k+1 in {{{n(n+1)/2}}} to see
what the Pk+1 is, for that is what we are going for, 
and if we have that beforehand, we'll know when we 
have arrived and the proof is finished.

Substituting k+1 for n in {{{n(n+1)/2}}}, we have

{{{(k+1)(k+1+1)/2}}} or {{{(k+1)(k+2)/2}}} or {{{(k^2+3k+2)/2}}}

Now that we know what Pk+1 is, we know where we're 
going, 

and we'll know we have arrived if and when we get 
{{{(k^2+3k+2)/2}}}. 

So now we can start the proof:

P1:  substitute n=1, {{{1=1(1+1)/2}}}, which is 
clearly true.

Assume Pk: {{{1 + 2 + 3 +""*""*""*""+ k}}}{{{""=""}}}{{{k(k+1)/2}}}

Add (k+1) to both sides:

{{{1 + 2 + 3 +""*""*""*""+k + (k+1)}}}{{{""=""}}}{{{k(k+1)/2 + (k+1)}}}

              {{{""=""}}}{{{(k^2+k)/2+2(k+1)/2}}}

              {{{""=""}}}{{{(k^2+k)/2+2(k+1)/2}}}

              {{{""=""}}}{{{(k^2+k)/2+(2k+2)/2}}}

              {{{""=""}}}{{{(k^2+k+2k+2)/2}}}

              {{{""=""}}}{{{(k^2+3k+2)/2}}}

and now we see that we get the same Pk+1 as the one
that we found in the beginning that we were going 
for.

So the first proof is finished. Since P1 is true, 
P1 proves P2, P2 proves P3, P3 proves P4, etc., etc., 
ad infinitum.

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Now we prove

Qn: {{{1^3+2^3+""*""*""*""+n^3}}}{{{""=""}}}{{{( (n(n+1))/2 )^2}}} 

As before, let's see what Qk+1 would be:

To do that, replace n by k+1 in {{{(n(n+1)/2)^2}}} to 
see what the Qk+1 is.  That is what we are going for, 
and as before, if we have that beforehand, we'll know 
when we have arrived and the proof is finished.

Substituting k+1 for n in {{{(n(n+1)/2)^2}}}, we have

{{{((k+1)(k+1+1)/2)^2}}} or {{{((k+1)(k+2)/2)^2}}} or {{{(k^2+3k+2)^2/4}}} or {{{(k^4+9k^2+4+6k^3+4k^2+12k)/4}}} 
or {{{(k^4+6k^3+13k^2+12k+4)/4}}}

Now that we know what Qk+1 is, we know where we're 
going, and we'll know we have arrived if and when we 
get {{{(k^4+6k^3+13k^2+12k+4)/4}}}. 

So now we can start the second proof:

Q1:  substitute n=1, {{{1=(1(1+1)/2)^2}}}, which is clearly 
true.

Assume Qk: {{{1^3+2^3+3^3+""*""*""*""+k^3}}}{{{""=""}}}{{{((k(k+1))/2)^2}}}

Add (k+1)³ to both sides:

{{{1^3+2^3+3^3+""*""*""*""+k^3+ (k+1)^3}}}{{{""=""}}}{{{(k(k+1)/2)^2 +(k+1)^3}}}

                  {{{""=""}}}{{{(k^2+k)^2/4+k^3+3k^2+3k+1}}}

                  {{{""=""}}}{{{(k^4+2k^3+k^2)/4+k^3+3k^2+3k+1}}}

                  {{{""=""}}}{{{(k^4+2k^3+k^2)/4+4(k^3+3k^2+3k+1)/4}}}

                  {{{""=""}}}{{{(k^4+2k^3+k^2)/4+4k^3+12k^2+12k+4)/4}}}

                  {{{""=""}}}{{{(k^4+2k^3+k^2+4k^3+12k^2+12k+4)/4}}}

                  {{{""=""}}}{{{(k^4+6k^3+13k^2+12k+4)/4}}}

and now we see that we get the same Qk+1 as the one 
that we found in the beginning that we were going for.

So the second proof is finished. Since Q1 is true, 
Q1 proves Q2, Q2 proves Q3, Q3 proves Q4, etc., etc., 
ad infinitum.
 
Edwin</pre></b></font>