Solving Math Induction: Prove 1^3 + 2^3 + 3^3 + 4^3 ... + n^3 = ((n^2 + n)/2)^3

  • Thread starter ruud
  • Start date
  • Tags
    Induction
In summary, the conversation is about proving a mathematical statement by induction. The speaker has shown their work for n=1 and n=2, and then assumes n=k and tries to prove for n=k+1. However, the result does not hold for n=2, and upon further calculation, it is clear that the statement is false. The speaker's friend points out this error, and the conversation ends with the realization that the statement is not true.
  • #1
ruud
14
0
I'm stuck and not sure what I've done wrong for this problem

Prove the following by mathematical induction:
1^3 + 2^3 + 3^3 + 4^3 ... + n^3 = ((n^2 + n)/2)^3

ok so I proved it for n = 1 and n = 2 then assume n = k
so ((k^2 + k)/2)^3

Then let's try to do n = k +1
so
((k+1)^2 + k +1)/2)^3 = ((k^2 + k)/2)^3 + (k + 1)^3
after expanding I get

(k^6+ 9k^5+ 33k^4+ 63k^3+ 66k^2+ 36k+ 8)/8
=
(k^6 +3k^5 + 3k^4 + 9k^3 + 24k^2 + 24k + 8)/8 + k^3 + 3k^2 + 3k +1

For some reason I think that I don't have to do all of this expanding. Can someone please tell me what I"m doing wrong or what I need to fix?
 
Mathematics news on Phys.org
  • #2
Nevermind I just heard from a fellow friend that this statement is false.
 
  • #3
Someone had to tell you that? Did you even try calculating a few numbers?

If n= 2, 12+ 22= 1+ 8= 9

[tex]\(\frac{n^2+n}{2}\)^3= \(\frac{4+2}{2}\)^3= 3^3= 27[/tex]

so for the very second number its not true.

If you had done even the slightest amount of work on this you would have seen:
1 13 = 1
2 13+23= 1+ 8= 9
3 13+23+33= 1+ 8+ 27= 36
4 13+23+33+43= 100.

Hmmm: 1, 9, 36, 100,... what does that make you think of? Not cubes certainly!
 

What is math induction?

Mathematical induction is a method of proof used to establish the validity of a statement for all natural numbers greater than or equal to a starting point (usually 0 or 1). It involves proving the statement for the starting point, and then showing that if it is true for a particular number, it must also be true for the next number.

How do you use math induction to prove a statement?

To prove a statement using mathematical induction, you must first establish the base case (usually n = 0 or n = 1) by showing that the statement holds true. Then, you assume that the statement is true for a particular number, and use this assumption to prove that it must also be true for the next number. This shows that if the statement is true for one number, it must be true for all subsequent numbers, thus proving the statement for all natural numbers greater than or equal to the starting point.

What is the statement being proved in this equation?

The statement being proved in this equation is that 1^3 + 2^3 + 3^3 + 4^3 ... + n^3 is equal to ((n^2 + n)/2)^3 for all natural numbers n.

What is the base case for this proof?

The base case for this proof is n = 1, where 1^3 is equal to ((1^2 + 1)/2)^3. This can be easily verified as true, as 1^3 = 1 and ((1^2 + 1)/2)^3 = ((1 + 1)/2)^3 = (2/2)^3 = 1^3 = 1.

How do you show that the statement is true for the next number?

To show that the statement is true for the next number, we assume that it is true for a particular number n, and then use this assumption to prove that it must also be true for n+1. In this case, we assume that 1^3 + 2^3 + 3^3 + 4^3 ... + n^3 is equal to ((n^2 + n)/2)^3, and then add (n+1)^3 to both sides. By simplifying and using algebraic manipulations, we can show that the statement holds true for n+1 as well.

Similar threads

Replies
13
Views
1K
  • General Math
Replies
8
Views
1K
Replies
5
Views
2K
Replies
1
Views
617
Replies
1
Views
717
Replies
2
Views
975
  • General Math
Replies
9
Views
188
Replies
17
Views
3K
  • General Math
Replies
1
Views
742
Back
Top