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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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- Thread starter A93
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- #1

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

- #2

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What did you try?? If you let us know where you're stuck, then we'll know how to help...

- #3

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my dumb butt got "lost" in the wanting to prove p(k+1) is true (if that even makes sense, lol)

- #4

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and where are you stuck in induction??

- #5

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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

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[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

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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?

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