SOLUTION: Identify the correct proof by induction or a counterexample to disprove the following statement. 1^3+2^3+3^3+...+n^3=n^2(n+1)^2/4

Algebra ->  Trigonometry-basics -> SOLUTION: Identify the correct proof by induction or a counterexample to disprove the following statement. 1^3+2^3+3^3+...+n^3=n^2(n+1)^2/4      Log On


   



Question 1035751: Identify the correct proof by induction or a counterexample to disprove the following statement.
1^3+2^3+3^3+...+n^3=n^2(n+1)^2/4

Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52794) About Me  (Show Source):
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bn%5E3%22%22=%22%22%28n%5E2%28n%2B1%29%5E2%29%2F4
induction proof:
 
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%5E2%28n%2B1%29%5E2%2F4 to see what 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 %28n%5E2%28n%2B1%29%5E2%29%2F4, we have

%28%28k%2B1%29%5E2%28%28k%2B1%29%5E%22%22%2B1%29%5E2%29%2F4 or 

%28%28k%2B1%29%5E2%28k%2B1%2B1%29%5E2%29%2F4 or

%28%28k%2B1%29%5E2%28k%2B2%29%5E2%29%2F4

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 %28%28k%2B1%29%5E2%28k%2B2%29%5E2%29%2F4. 

So now we can start the proof:

P1:  substitute n=1, 
1%5E3%22%22=%22%22%281%5E2%281%2B1%29%5E2%29%2F4%22%22=%22%22%281%282%29%5E2%29%2F4%22%22=%22%224%2F4%22%22=%22%221,
which is true.

Assume Pk: 1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%22%22=%22%22%28k%5E2%28k%2B1%29%5E2%29%2F4

Add (k+1)³ to both sides:

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2Bred%28%28k%2B1%29%5E3%29%22%22=%22%22%28k%5E2%28k%2B1%29%5E2%29%2F4%2Bred%28%28k%2B1%29%5E3%29
 
Get an LCD of 4

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28k%5E2%28k%2B1%29%5E2%29%2F4%2B4%28k%2B1%29%5E3%2F4

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28k%5E2%28k%2B1%29%5E2%2B4%28k%2B1%29%5E3%29%2F4

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

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28%28k%2B1%29%5E2%28k%5E2%2B4%28k%2B1%29%29%29%2F4 

Distribute on right:

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28%28k%2B1%29%5E2%28k%5E2%2B4k%2B4%29%29%2F4

Factor on right:

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28%28k%2B1%29%5E2%28k%5E2%2B4k%2B4%29%29%2F4

1%5E3%2B2%5E3%2B3%5E3%2B%22...%22%2Bk%5E3%2B%28k%2B1%29%5E3%22%22=%22%22%28%28k%2B1%29%5E2%28k%2B2%29%5E2%29%2F4

and now we see that we have gotten Pk+1 that we found in the 
beginning that we were going for.

So the proof is finished.

So since P1 is true, P1 proves P2, P2 proves P3, P3 proves P4,
etc., etc., ad infinitum.

Edwin