harder inequality proves (1 Viewer)

[jmromeo]

Can any body show me how to prove this:

1^3+2^3+3^3+...+n^3 <= n^4/3, using the area under the curve y=x^3 from x=1 to x=n. This question appears on the Patel's textbook (second edition) page 314.

 

[KFunk]

Are you sure you wrote the question out correctly? If you take an example:

1³ + 2³ + 3³ + 4³ + 5³ = 225

I figured there were two ways to read what you wrote:

(n⁴)/3 = 5⁴/3 = 208 1/3

or n^4/3 = 5^4/3 = 8.549...

Either way, in both cases 1³ + 2 ³ + ... + n³ &ge; n^3/4

Check the question to see if you left anything out or let me know if I've interpreted what you've said incorrectly.

 

[jmromeo]

Harder inequality prove

Thanks for replaying...! I doubled check the question and it goes like this:

1^3+2^3+3^3+...+n^3 <= (n^4)/3, but now I have the feeling that the

question is wrong. I think you are wright.

 

[who_loves_maths]

^ the question implies the need to integrate. so i don't see how the book got (n^4)/3 at all.

in addition, if the inequality sign is in the right direction, then the question should have asked you to perhaps consider the area under the curve from 1 to (n+1), and not just n.

by doing that, at least you would arrive at the inequality:

[(n+1)^4 - 1]/4 = n(n+2)(n^2 + 2n+ 2)/4 >= 1^3 + 2^3 + ... + n^3

 

[jmromeo]

Harder inequality prove

Thanks guys!...I think the book is wrong!...I wish you do well in your exams!