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## What do these new symbols mean...?I can't start this without knowing

Oh if f = 1, then P = 1.

Can I jsut state that or should I write up more on nCr?

Mentor  That can't be right. P(x) doesn't depend on f(x) at all. Try again.

In the summation I get  $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$

Oh wait, there are n - k + 1 terms...shoot I forgot how to do this. I am guessing that means the sum sums to 1.

Mentor

Quote by flyingpig 					 				 				So I must find  $$P = \binom{n}{k} x^k (1 - x)^{n-k} = 1$$
Why? Where did you get this from?

Quote by vela 					 				 				Why? Where did you get this from?
Yeah scratch that I aws still thinking of multiplying and stuff. Forgot to get rid of it.

In  $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$  It can only be 1 when k = 0, and n = 0...  Is that what I Have to show?  This is due tomorrow and I m still scratching my head

In the picture, f(x) was shown to be $$f(\frac{k}{n})$$ does this imply that k = n since f(x) = 1?  If so then  $$\frac{n!}{n!(0!)} x^n (1-x)^0 = x^n$$  Darn something is still off.

Mentor  Do you understand sigma notation?

Quote by vela 					 				 				Do you understand sigma notation?
It means sum

EDIT: oh wait, are you referring when I said that n = k? Oh okay, so the sum is $$\sum_{n=0}^{n} \{whatever is here} = 1$$?

EDITING..not actually too sure of the property above. Looking through my old calc text

Mentor  I'm asking you what is $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$ shorthand for? In other words, if you write the sum out, what do you get? I'm asking because you're making a bunch of guesses that make no sense if you understand what the notation means.

$$\binom{n}{k} x^k (1 - x)^{n-k} = P$$  $$\sum_{k=0}^{n} P$$  I am sorry for being so slow.   wEDIT:writing out the sum...

I wrote the first three terms $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k} = (1-x)^n + \frac{n!}{(n-1)!}x(1-x)^{n-1} + \frac{n!}{2!(n-2)!}x^2 (1-x)^{n-2}...$$

Mentor  Ok, good. So you see how k isn't a variable you can really mess with, right? It takes on the values 0 to n just to generate the terms in the sum. In fact, when you expand the sum out, there is no k appearing anymore because it was just a dummy variable. So you can't do stuff like assume restrict k to just one value to try get the result you want.  Similarly, you're asked to evaluate the sum for any value of n, so you can't set n to any one value.  Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)n?

$$(a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k$$    By observations I took  x = b  1 - x = a  So that a + b = 1  1^n = 1 forever, so that completes the problem!!! YES THANK YOU VELA, Kurt, and micro and all who helped  *blows kisses*

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