Summing Factorials: Solving the Homework Statement

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    Factorial Sum
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Homework Help Overview

The problem involves summing a series related to factorials, specifically the expression \(\sum_{n=0}^{100} \frac{1}{n!(100-n)!}\). Participants are exploring connections to binomial coefficients and the binomial theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the given sum and binomial coefficients, questioning how to interpret and manipulate the expression. There are attempts to connect the sum to the expansion of \((1+1)^{100}\) and the implications of this expansion.

Discussion Status

The discussion is ongoing, with participants expressing confusion and seeking clarification on the connections between the factorial sum and binomial coefficients. Some guidance has been offered regarding the binomial theorem, but no consensus has been reached on the next steps.

Contextual Notes

Participants mention a deadline for the homework, indicating a sense of urgency. There are also hints of uncertainty regarding the application of the binomial theorem and the expansion of \((1+1)^{100}\).

dystplan
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Homework Statement


[itex]\sum_{n=0}^{100} 1/n!(100-n)![/itex]

The Attempt at a Solution


Other then obvious attempts to make sense of the equation's incremental and decrements divisor, I can't figure out where to start with this question. Some assistance would be greatly appreciated.
 
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It looks like it's closely related to the sum of binomial coefficients. What's the sum of C(100,n) for n from 0 to 100? Is that enough of a hint?
 
C(100,n) being = 100!/n!(100-n)! ? hmmm, unfortunately I don't see where that's going =/

Man this one is throwing me for a loop, only question I haven't managed and due tomorrow.
 
dystplan said:
C(100,n) being = 100!/n!(100-n)! ? hmmm, unfortunately I don't see where that's going =/

Man this one is throwing me for a loop, only question I haven't managed and due tomorrow.

Yes, that's C(100,n). There is a simple formula for the sum of the binomial coefficients. It's related to the value of (1+1)^100. Don't know it? Expand (1+1)^100 using the binomial theorem.
 
well; thanks. But I'm still just totally lost. (1+1)^100 is a massive equation when expanded.
 
Unless... x/y = 0 or 1 in the binomial theorem? That would make it easy.

Makes it = 1/100! ?
 
dystplan said:
well; thanks. But I'm still just totally lost. (1+1)^100 is a massive equation when expanded.

Oh come on, (1+1)^100=2^100. That's an easy enough number to write down. Now what does that have to do with the sum of the binomial coefficients C(100,n)?? C(100,0)+C(100,1)+...+C(100,100). I'm not asking you to evaluate each one. Just think about it.
 
wait;

2^100/100!
 
wait;

2^100/100!
 
  • #10
dystplan said:
wait;

2^100/100!

You aren't just guessing, I hope.
 

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