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Trigonometry 13 Online
OpenStudy (anonymous):

Use mathematical induction to show that this statement is true. I will draw a diagram. I appreciate any help!

OpenStudy (anonymous):

\[\sum_{i=1}^{n}i^3 = [n(n+1)/2]^2\]

OpenStudy (tkhunny):

Does it work for n = 1?

OpenStudy (anonymous):

I do believe it does. I got that it does. I am not sure how to prove the second part where you have to prove if it is k+1 which is the second half of mathematical induction

OpenStudy (tkhunny):

You believe? Does it or doesn't it?

OpenStudy (anonymous):

It does

OpenStudy (tkhunny):

Perfect. Proceed with confidence. If it works for a value, k, we have: \(\sum\limits_{i=1}^{k} i^{3} = \left(\dfrac{k(k+1)}{2}\right)^{2}\). Can we use this fact to demonstrate that: \(\sum\limits_{i=1}^{k} i^{3} + (k+1)^{3} = \left(\dfrac{(k+1)(k+2)}{2}\right)^{2}\)

OpenStudy (anonymous):

Yes we can. That makes sense. The k+1 always throws me off. Thank you for the motivation on confidence. The class I am has really lowered it. Now all I need to do is factor the second equation our farther and it should be perfect

OpenStudy (tkhunny):

It's a little bit of algebra. You can do this!

OpenStudy (anonymous):

I pretty sure I got it! My equations matched!

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