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

[flyingpig]

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

 

[vela]

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.

 

[flyingpig]

\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...

 

[flyingpig]

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}...

 

[vela]

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 anyone value.

Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)ⁿ?

 

[flyingpig]

(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*