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# The Proof is Trivial! watch

1. (Original post by Mladenov)
Hint (Problem 23)

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Write , so . Now for each , consider , , and . Write , and note that takes each value from exactly once. We are now able to evaluate each of these three sums. Then sum over to find .

There should be a more concise solution.
There is

There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods
2. Subbing post (and admission of spending a couple of hours here instead of doing revision...)
3. (Original post by shamika)
There is

There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods
It's an interesting problem, and it looks like:

and for:

Where is the nth pentatopic number.
4. (Original post by Noble.)
It's an interesting problem, and it looks like:

and for:

Where is the nth pentatopic number.
So for the original sum...
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?
Can you prove it?
5. Solution 29

By Van der Waerden's theorem must contain arbitrarily long arithmetic progressions hence there are infinitely many finite sequences for which with . By fixing we have as desired. There are infinitely many such sequences hence there are infinitely many ordered pairs
6. Problem 30*

More limits & integrals?

Evaluate
7. (Original post by metaltron)
...
You have integrated incorrectly,

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Edit: by the way, this is going to seem pedantic but the notation is meaningless (use an equal sign)
8. (Original post by Lord of the Flies)
You have integrated incorrectly,

Spoiler:
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Edit: by the way, this is going to seem pedantic but the notation is meaningless (use an equal sign)
Thanks. Got rid of it now as it was all completely wrong then! Might look at it again tomorrow when I'm more awake.

Spoiler:
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Thinking about it is the answer 1 as when 0 < x < 1, 1 + x^n = 1 as n tends to infinity. So the area under the curve becomes a rectangle area 1, which when raised to the power of n is still 1?
9. (Original post by Indeterminate)
...
Check that expansion - it isn't valid on the interval!
10. (Original post by Lord of the Flies)
Check that expansion - it isn't valid on the interval!

Indeed

I was wondering why it seemed so easy. I'll try something different
11. Solution 30

Using the binomial expansion and integrating, we find that:

12. (Original post by und)
Solution 30

-snip-

Lol, way to trip right in front of the finish line..

EDIT: Right, ignore this then.
13. (Original post by aznkid66)
Lol, way to trip right in front of the finish line..
Blithering idiot.

Lol, I got warning points for this.
14. Problem 31*

Find all real numbers x, y and z which satisfy the simultaneous equations , and .
15. (Original post by und)
Problem 31

Find all real numbers x, y and z which satisfy the simultaneous equations , and .
Solution 31

Adding the three equations:

Completing the square:

Hence
16. Problem 32*

Find all positive integers such that and are both perfect squares.
17. (Original post by und)
...
(Original post by Upper Echelons)
...
I thought und answered his own question there for a second

Solution 32
Solution 32
Let and where Isolating 'n' and equating the two, we get that

Running through the cases where two integers multiply together to make 819...

Just a horrendous case bash now ;_;

Bearing in mind that a, b are integers, we have to find the integer solutions. This only occurs when the difference of the two factor is a multiple of 10 because it reduces down to which only works for the pairs of

Solving simultaneously, for we get the solutions If we sub these back into and ;_;

and

We only take the integer solutions, so we get and

I'm sure there's a much more elegant solution than this disgusting case bash
18. Problem 33

Determine
19. (Original post by Benjy100)
Problem 33

Determine
Surely this has asymptotes in the interval you've given?
20. Solution 33

Method 1:

Letting gives:

Method 2:

(taking the principal branch - I'm sure that can be justified)

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