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What do these new symbols mean...?I can't start this without knowing 
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#55
Sep1911, 04:20 PM

P: 2,568

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.



#56
Sep1911, 04:21 PM

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#57
Sep1911, 04:23 PM

P: 2,568




#58
Sep1911, 09:13 PM

P: 2,568

In
[tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1  x)^{nk}[/tex] 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 


#59
Sep1911, 09:16 PM

P: 2,568

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


#60
Sep1911, 09:24 PM

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Do you understand sigma notation?



#61
Sep1911, 09:24 PM

P: 2,568

EDIT: oh wait, are you referring when I said that n = k? Oh okay, so the sum is [tex]\sum_{n=0}^{n} \{whatever is here} = 1[/tex]? EDITING..not actually too sure of the property above. Looking through my old calc text 


#62
Sep1911, 09:28 PM

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I'm asking you what is [tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1  x)^{nk}[/tex] 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.



#63
Sep1911, 09:31 PM

P: 2,568

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


#64
Sep1911, 09:35 PM

P: 2,568

I wrote the first three terms
[tex]\sum_{k=0}^{n} \binom{n}{k} x^k (1  x)^{nk} = (1x)^n + \frac{n!}{(n1)!}x(1x)^{n1} + \frac{n!}{2!(n2)!}x^2 (1x)^{n2}...[/tex] 


#65
Sep1911, 09:41 PM

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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}? 


#66
Sep1911, 09:48 PM

P: 2,568

[tex](a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{nk} b^k[/tex]
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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