Sum of n^3 = (sum of n)^2 induction proof

In summary: So at some point your inequality must hold.In summary, the conversation discusses proving the statement 1^3 + 2^3 +...+n^3=(1+2+...+n)^2 for natural numbers n using induction. The participants suggest adding (k+1)^3 to the left side and breaking down the right side into two parts, after k+1 and after k, to prove the equality. They also discuss another problem involving n^2<=2n and suggest using proof by induction or contradiction to show its validity.
  • #1
cnwilson2010
8
0

Homework Statement


Prove:
1^3 + 2^3 +...+n^3=(1+2+...+n)^2
n=Natural number

Homework Equations





The Attempt at a Solution


Using induction -
n=1 obvious

Assume for n=k equation is true.

Show for k+1.

I have that the right side prior to k+1 is (k^2(k+1)^2)/4
After k+1 I have (k^4+6k^3+13k^2+12k+4)/4,
but I can't figure out the trick to get the form above equal to the left +(k+1)^3.

Any ideas?
Thanks.
 
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  • #2
welcome to pf!

hi cnwilson2010! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

erm :redface: … a2 - b2 = … ? :wink:
 
  • #3
cnwilson2010 said:

Homework Statement


Prove:
1^3 + 2^3 +...+n^3=(1+2+...+n)^2
n=Natural number

Homework Equations





The Attempt at a Solution


Using induction -
n=1 obvious

Assume for n=k equation is true.

Show for k+1.

I have that the right side prior to k+1 is (k^2(k+1)^2)/4
After k+1 I have (k^4+6k^3+13k^2+12k+4)/4,
but I can't figure out the trick to get the form above equal to the left +(k+1)^3.

Any ideas?
Thanks.

On the left side you've added (k+1)^3. On the right side you've added the results after k+1 minus the results after k. The thing you want to prove equal to (k+1)^3 is (k+1)^2*(k+2)^2/4-k^2*(k+1)^2/4.
 
  • #4
Thank you for your replies they were very helpful. Now I'm working on n2<=2n.

For what values of Natural numbers is this statement true and prove by induction. Obviously, 1 and 2 are true, now I'm thinking proof by contradiction and the idea of the existence of a least member of the set. Does this sound reasonable or way too much work for the results?
 
  • #5
Seems begging for proof by induction, you will have to show that [itex]2k+1<k^{2}[/itex] at some point.

Or failing that, you could note that the two sequences [itex]a_{n}=n^{2},b_{n}=2^{n}[/itex] are both monotonically increasing sequences and that one increases faster that the other.
 

1. How can I prove that the sum of the cubes of the first n natural numbers is equal to the square of the sum using mathematical induction?

The proof using mathematical induction involves showing that the statement is true for the base case (n=1) and then showing that if the statement is true for some value of n, it is also true for the next value (n+1). By doing this, we can prove that the statement is true for all natural numbers.

2. What is the significance of proving this statement using induction?

Proving this statement using mathematical induction is significant because it is a powerful and widely used technique in mathematics for proving statements about natural numbers. It allows us to prove a statement for infinitely many cases by only having to show it for a few specific cases.

3. Can you explain why the sum of the cubes of the first n natural numbers is equal to the square of the sum?

This statement can be proven using algebraic manipulation. By expanding the square of the sum, we get the sum of the first n natural numbers squared. Then, by using the formula for the sum of the first n natural numbers and simplifying, we can see that the two expressions are equivalent.

4. Does this proof hold true for all sets of numbers?

No, this proof only holds true for the set of natural numbers. It does not hold true for other sets of numbers such as integers or real numbers.

5. Are there any other methods for proving this statement?

Yes, there are other methods for proving this statement such as using geometric or combinatorial arguments. However, mathematical induction is the most commonly used and efficient method for proving statements about natural numbers.

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