## 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*