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Converting Sigma notation... 
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#1
Jun2811, 01:54 AM

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1. The problem statement, all variables and given/known data
(Not a homework question) Hi!! I have been encountering problems in Binomial Theorem which includes converting the sigma notation. Like [tex]\sum_{k=0}^n \frac{n!}{(nk)!k!} a^kb^{nk}=(a+b)^n[/tex] I got many questions in my exam of this type with four options. One of them was: [tex]\sum_{k=1}^{n} {}^nC_k.3^k[/tex] I substituted the value of n and was able to figure out the correct option. But as i said there were many questions, so it took a lot of time. Is there any easier way to do that? 2. Relevant equations 3. The attempt at a solution 


#2
Jun2811, 03:54 AM

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[tex]\sum_{k=0}^n \frac{n!}{(nk)!k!}a^kb^{nk}=(a+b)^n[/tex] So this would assume that a=b=1. I'm not sure if this was a question or you just incorrectly wrote down the binomial expansion though... 


#3
Jun2811, 04:32 AM

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(Would you please tell me how to make the "n" before "C" in Superscript?) I hope you get it now. 


#4
Jun2811, 07:00 AM

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Converting Sigma notation...
[tex]\sum_{k=1}^n ^nC_k.3^k[/tex] to create a superscript in [itex]\LaTeX[/itex] just use ^ (and add {} for multiple characters) before it Notice that we can see a=3, but b isn't present. In fact, it's just hidden as b=1 because [tex]\sum_{k=1}^n ^nC_k.3^k1^{nk}[/tex] is exactly the same thing. So our final answer would be [tex]\sum_{k=0}^n ^nC_k.3^k^nC_03^0=(3+1)^n1=4^n1[/tex] EDIT: and it seems that latex has changed, once again... Man I'm getting annoyed with it. Give me a second and I'll figure it out. 


#5
Jun2811, 09:39 AM

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#6
Jun2811, 10:13 AM

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[tex]^nC_03^0[/tex] 


#7
Jun2811, 12:31 PM

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Hi PranavArora!
Seeing that Mentallic and vela are not around, I'll answer your question. It's part of the sigma notation and its implications. In particular this is about the boundaries of the sum, which in your case is starting with k=1. What you have is: [tex]\begin{eqnarray} \sum_{k=1}^n {^n}C_k \cdot 3^k &=&{^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + \dotsb + {^n}C_n \cdot 3^n\\ &=&({^n}C_0 \cdot 3^0 + {^n}C_1 \cdot 3^1 + {^n}C_2 \cdot 3^2 + \dotsb + {^n}C_n \cdot 3^n)  {^n}C_0 \cdot 3^0 \\ &=&(\sum_{k=0}^n {^n}C_k \cdot 3^k)  {^n}C_0 \cdot 3^0 \end{eqnarray}[/tex] The way to change the boundaries is always the same. Your write out the sum in its terms, change what you want to change, and change it back again into sigma notation. Note that the sigma notation is only a shorthand notation. Don't think of it as something magical that has its own rules  it hasn't. It's just shorthand. 


#8
Jun2811, 12:38 PM

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But what i would do if a question appears like this: [tex]\sum^n_{k=0} (2k+1) {}^nC_k[/tex]. 


#9
Jun2811, 01:18 PM

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The method I know is to define a function of x and integrate it. That is: [tex]s(x) = \sum^n_{k=0} {}^nC_k (2k+1) x^{2k}[/tex] The result you're looking for in this case is s(1). If you integrate it, you should find a form that looks more like your previous problem. You can rewrite that without the sigma and binomium. Afterward you differentiate again. And finally you fill in the value 1. Care to try? 


#10
Jun2811, 08:32 PM

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#11
Jun2811, 11:08 PM

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#12
Jun2811, 11:20 PM

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#13
Jun2811, 11:22 PM

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[tex]\frac{2(x^{2k+1})}{2k+1}[/tex] 


#14
Jun2811, 11:24 PM

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#15
Jun2811, 11:38 PM

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#16
Jun2911, 12:29 AM

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Here's the formula: [tex](1+x)^n=C_0+C_1x+C_2x^2+........+C_kx^k+.......+C_nx^n...[/tex] Integrating the above equation with respect to x between limits 0 to 1, (I don't understand what it is ) we get, [tex](\frac {(1+x)^{n+1}}{n+1})_0^1=(C_0x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+.....C_ n\frac{x^{n+1}}{n+1})_0^1[/tex] [tex]=C_0+\frac{C_1}{2}+\frac{C_2}{3}.....\frac{C_n}{n+1}=\frac{2^{2n+1}1}{n+1}[/tex] 


#17
Jun2911, 12:57 AM

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I have only one formula which involves integration within limits. But i only know how to integrate without limits. I have never solved questions which involve integration within limits. Here's the formula: [tex](1+x)^n=C_0+C_1x+C_2x^2+........+C_kx^k+.......+C_nx^n...[/tex][/quote] If that is the formula, then when evaluating this at 1, we would have [tex](1+1)^n=2^n=C_0+C_1+C_2+...+C_k+...+C_n[/tex] correct? and evaluating it at 0 [tex](1+0)^n=1^n=1=C_0+C_1\cdot 0+C_2\cdot 0+...=1[/tex] 


#18
Jun2911, 01:25 AM

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I like Serena said that define a function of x and integrate it but i still don't get how he got x^{2k}? 


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