Prove that for each n in N, 1^3+2^3+ +n^3=[n(n+1)/2]^2

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In summary, the steps for proving the summation formula for each n in N, 1^3+2^3+...+n^3=[n(n+1)/2]^2 using induction are: 1) showing the base case for k = 1, 2) creating an induction hypothesis by assuming the kth case holds, and 3) proving that the kth case implies the (k+1)th case.
  • #1
A93
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prove that for each n in N, 1^3+2^3+...+n^3=[n(n+1)/2]^2
 
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  • #2
What did you try?? If you let us know where you're stuck, then we'll know how to help...
 
  • #3
induction.
my dumb butt got "lost" in the wanting to prove p(k+1) is true (if that even makes sense, lol)
 
  • #4
and where are you stuck in induction??
 
  • #5
the proving of p(k+1) is true.
basically the area where for each k>=1, if P(k) is true, then p(k+1) is true...basically the induction part, lol
 
  • #6
Yes, so to show that p(k+1) is true, you need to prove that

[tex]1^3+2^3+...+k^3+(k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2[/tex]

Now, what happens if you appy "p(k) is true" on that??
 
  • #7
A93 said:
induction.
my dumb butt got "lost" in the wanting to prove p(k+1) is true (if that even makes sense, lol)

Induction has a few steps. Let's see if this clarifies them a bit,

1) Base Case: Show that your summation formula works for k = 1 case (which is probably easiest here lol)

2) Induction Case: Create an induction hypothesis. For this case, you assume that the kth case holds. In other words,

[itex]\sum_{k=1}^{n}k^3=\left (\frac{n(n+1)}{2} \right )^2[/itex]

is true. Now, show that the kth case implies the (k+1)th case. How do you think you can do this?
 
Last edited:

1. What is the equation being proven?

The equation being proven is 1^3+2^3+ +n^3=[n(n+1)/2]^2.

2. What does N in the equation represent?

N represents the set of natural numbers, which are counting numbers starting from 1 (i.e. 1, 2, 3, ...).

3. What is the purpose of proving this equation?

The purpose of proving this equation is to demonstrate a mathematical relationship between the sum of cubes of natural numbers and the square of the sum of the same natural numbers.

4. How can this equation be proven?

This equation can be proven using mathematical induction, which involves showing that the equation holds for a base case (n=1) and then showing that if it holds for any arbitrary number k, it also holds for the next number (k+1).

5. What implications does this equation have in mathematics?

This equation has implications in various fields of mathematics, including number theory, algebra, and calculus. It also has practical applications in solving problems related to sums of consecutive cubes and squares.

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