Max0815

What goes into a black hole stays in a black hole according to modern physics.

Perhaps somebody comfirm my answers?

K.

Yea.

Oh, Is it like messages?

Hi. What is a chat room?

@sempaiscuba ok thanks.

@sempaiscuba ok...I see. thanks, Also, is my view of goliad correct? The Texas revolution made it seem like Mexico has a super bad military, in which it coudnt ever crush rebelions that were fairly small. Is this actually trye?

@sempaiscuba Hmmm...I see :P but the defeat.. wouldn't that scare many people, which would be more than the number of people who deemed revenge? Santa Anna executed all who surrendered, and I believe that he did it to scare people. I reason that if I got out there to fight, I would either be killed fighting, or killed after surrendering. It seems too risky.

@sempaiscuba Shakespeare is not a reliable source. He was a lay writer and oftenly exaggerated everything

@sempaiscuba however not all of them fought to death heroicly, and according to some sources, David C. surrendered...

@sempaiscuba how was that a "heroic" defeat?

@sempaiscuba yes. "News of the Alamo's fall had the opposite effect, however, and men flocked to join Houston's army.[150] The New York Post editorialized that "had [Santa Anna] treated the vanquished with moderation and generosity, it would have been difficult if not impossible to awaken that general sympathy for the people of Texas which now impels so many adventurous and ardent spirits to throng to the aid of their brethren"." However, i do not get why the defeat made men join the army.

I don't see how it can be answers from a single link, furthermore this question is not a "too basic" question. What can i search up? I mean, do i look up "list of texas independence events"???

Why downvote???

What is revenge downvoting?

Oh....i dont think it is really meant to be a combinatorial proof.

And again, I want to thank you all very much! :)

Ok. Thanks! Ok, so it doesn't need to be fixed or anything? :) ...I don't get what you mean by that you cant think of any..i thought you just wrote a proof in this post?

@TB it was long.. We know a generating function whose coefficients are of the form $\binom n2$: namely, $$\frac 1{(1-x)^3} = \binom 22 + \binom 32x + \binom 42x^2 + \cdots.$$ Substituting $x=\frac 14$ in this expansion, we get $$\frac 1{\left(1-\frac 14\right)^3} = \binom 22 + \frac{\binom 32}4 + \frac{\binom 42}{16} + \cdots.$$ Dividing by $16$ gives $$\frac 1{16}\cdot \frac 1{\left(1-\frac 14\right)^3} = \frac{\binom 22}{16} + \frac{\binom 32}{64} + \frac{\binom 42}{256} + \cdots.$$ The right side is the sum we wished to determine, and the left side simplifies to $\boxed{\frac 4{27}}$.

@Theo Bendit Its ok. :) They say if you don't give up, you learn, and that's exactly what happened today :) I actually solved it :) through the 4 hints (I clicked on edit and looked at them :P). Thanks to everyone who helped! Your answers and appreciated!!!!!!

@anyone, especially to Theo Bendit. My solution is below. Is this a good approach and solution? Can there be any improvements???

@Guido A. its ok, I think I still can solve the problem... I think. (ahh I cant do another @) So Theo Bendit weren't all four hints along the way to solve the problem??? Why did you take those away?

@Theo Bendit thanks for your hints, I will play around for a while, and see if I get to any avail. Thanks.

@Guido A. your convergence was quite useful into moving deeper into the problem, but as I went to dinner, it was deleted and I couldn't continue. However, maybe this can help me too. Thanks for your help though :)

I don't know why the previous answer was deleted. Yes, may you please expand that?

Hold on...let me try again.