Math Q&A Game

StatusX

That looks correct, but I think there's a simpler way.

StatusX

Ok, forget that one. Here's a little one I came up with screwing around during a boring class today. Find:

[tex]\sum_{n=0}^{\infty} {\left( \begin{array}{c} n+k \\ k \end{array} \right)}^{-1} [/tex]

EDIT: Ok, Latex is working now.

NateTG

That's the sum of the reciprocals of the binomial coefficients?

StatusX

Is this too easy or too hard? Here's a clue: (click to get it. the white text was showing up, so I thought this would be a good way to hide it.)

[tex] \telescoping \series [/tex]

Moo Of Doom

White text might still be visible, but color #e9e9e9 text is impossible to read.

StatusX

Well I'll give the answer and let someone else go if they want:

[tex]\sum_{n=0}^N \left( \begin{array}{c} n + k \\ k \end{array} \right)^{-1} = \sum_{n=0}^N \frac{n! k!}{(n+k)!}[/tex]

[tex] = k! \sum_{n=0}^N \frac{1}{(n+k)(n+k-1)...(n+1)}= k! \sum_{n=0}^N \frac{1}{(n+k)(n+1)}\frac{1}{(n+k-1)...(n+2)}[/tex]

[tex]=k! \sum_{n=0}^N \frac{1}{k-1}\left(\frac{1}{n+1} - \frac{1}{n+k} \right) \frac{1}{(n+k-1)...(n+2)}=
\frac{k!}{k-1} \sum_{n=0}^N \left(\frac{1}{(n+k-1)...(n+2)(n+1)} - \frac{1}{(n+k)(n+k-1)...(n+2)} \right) [/tex]

But this is a telescoping series, so we get:

[tex]\sum_{n=0}^N \left( \begin{array}{c} n + k \\ k \end{array} \right)^{-1} = \frac{k!}{k-1} \left( \frac{1}{(k-1)...(2)(1)} - \frac{1}{(N+k)(N+k-1)...(N+2)} \right) = \frac{k}{k-1} - \frac{k!}{(k-1)(N+k)...(N+2)}[/tex]

Which goes to k/(k-1) as N goes to infinity.

darwid

What was the elegant solution for this one ? I tried for 2 days and couldn't find it ;)

StatusX

Woa, sorry, missed your post. And, yea, sorry again, I didn't have one in mind, I just assumed there was one. I still think there is, but I haven't found it yet either. I'll get back to you (if you still exist).

whatta

I take it there is no on-going puzzle here, so I thought I'll post something. Solve this proportion: 11/2 = 3/10, 10/8 = ?/10.

StatusX

?=6 (mod 13)

whatta

well it could be, but I had something completely different in mind (hint: / was not supposed to mean division)

shamsoddin

I think it was better that you specify more equations not just one.(for example two of them).But anyway I think :
?=8

whatta

yep, that was it! more equations for those who did not figured yet: 1001/2 =11/8, 112/3 = 15/9, etc.

StatusX

Ok, I see. Yea, more equations would have been a good idea to eliminate other possibilities, since 11/2 = 3/10 is also true in the field Z/13Z (ie, mod 13):

11/2=24/2=12
3/10=120/10=12

Your turn shamsoddin.

shamsoddin

This is my quastion :
Suppose function F is continuous in [2,4] and differntiatable in (2,4).We have
F(2)=2 and F(4)=4.Prove a point C exists that tangent to the corresponding curve in this point C passes the origin i.e this tangent line includes the origin.

whatta

cant we say, if it does not exist, all angles have to be either less than or greater than 45 and so integrating it from (2,2) onwards wouldn't ever come to (4,4)?

shamsoddin

It is not a persist proof and in mathematics persist ones are needed.So I give you some hints :You should define a function and then use Roll's Theorem to solve this problem.
good luck !