Question 1035751
{{{1^3+2^3+3^3+"..."+n^3}}}{{{""=""}}}{{{(n^2(n+1)^2)/4}}}
<pre>
induction proof:
 
First let's see what P<sub>k+1</sub> would be:
[That's always the first thing to do.  Before you start an induction
proof,  you should calculate P<sub>k+1</sub> to see where you're headed]:
To do that, replace n by k+1 in {{{n^2(n+1)^2/4}}} to see what P<sub>k+1</sub> 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^2(n+1)^2)/4}}}, we have

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

{{{((k+1)^2(k+1+1)^2)/4}}} or

{{{((k+1)^2(k+2)^2)/4}}}

Now that we know what P<sub>k+1</sub> is, we know where we're going, 
and we'll know we have arrived if and when we get {{{((k+1)^2(k+2)^2)/4}}}. 

So now we can start the proof:

P<sub>1</sub>:  substitute n=1, 
{{{1^3}}}{{{""=""}}}{{{(1^2(1+1)^2)/4}}}{{{""=""}}}{{{(1(2)^2)/4}}}{{{""=""}}}{{{4/4}}}{{{""=""}}}{{{1}}},
which is true.

Assume P<sub>k</sub>: {{{1^3+2^3+3^3+"..."+k^3}}}{{{""=""}}}{{{(k^2(k+1)^2)/4}}}

Add (k+1)³ to both sides:

{{{1^3+2^3+3^3+"..."+k^3+red((k+1)^3)}}}{{{""=""}}}{{{(k^2(k+1)^2)/4+red((k+1)^3)}}}
 
Get an LCD of 4

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

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

Factor out (k+1)² in the numerator on the right:

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

Distribute on right:

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

Factor on right:

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

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

and now we see that we have gotten P<sub>k+1</sub> that we found in the 
beginning that we were going for.

So the proof is finished.

So since P<sub>1</sub> is true, P<sub>1</sub> proves P<sub>2</sub>, P<sub>2</sub> proves P<sub>3</sub>, P<sub>3</sub> proves P<sub>4</sub>,
etc., etc., ad infinitum.

Edwin</pre>