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lilsis76 Group Title

how can i find the 100th term in this problem: (1 + x)^100

  • 2 years ago
  • 2 years ago

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  1. surdawi Group Title
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    is x the number of terms?

    • 2 years ago
  2. lilsis76 Group Title
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    |dw:1353281221368:dw| this is how the problem is shown. says to find the 100th term. but I never understood :/

    • 2 years ago
  3. surdawi Group Title
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    ok i think that u have to plug 100 for x 101^100 which is lots of numbers

    • 2 years ago
  4. lilsis76 Group Title
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    lol okay so it would be|dw:1353281421656:dw|

    • 2 years ago
  5. zepdrix Group Title
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    Hmm this is a 100th degree binomial. When you expand it out, you should get 101 terms. We want the 100th term that is produced. Are you familiar with pascal's triangle? :D It provides a nice pattern for finding a certain term in binomial expansion.

    • 2 years ago
  6. lilsis76 Group Title
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    yes i know about the pascal triangle.

    • 2 years ago
  7. lilsis76 Group Title
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    but would i really have to write it all out?!

    • 2 years ago
  8. zepdrix Group Title
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    no no no course not :D but we should be able to see a pattern and predict the 100th term hmm

    • 2 years ago
  9. zepdrix Group Title
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    |dw:1353281609783:dw| See how pascal provides the coefficients for each term? With a binomial, they expand in such a way that each term is the product of both terms from the binomial, with the power on the first term starting at 3 (or whatever the degree of the binomial was) and counting down, while the other term starts at 0 and counts up.

    • 2 years ago
  10. zepdrix Group Title
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    There's a formula for generalizing this, but I think that will be alil more confusing than just recognizing the pattern :D

    • 2 years ago
  11. lilsis76 Group Title
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    okay, i see how that works

    • 2 years ago
  12. zepdrix Group Title
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    \[\large (1+x)^{100}=\]\[\large 1^{100}x^0+100*1^{99}x^1+...+100*1^1x^{99}+1^0x^{100}\]

    • 2 years ago
  13. zepdrix Group Title
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    There are a bunch of terms in the middle, we wouldn't be able to calculate their coefficients unless we went over the formula for generating them. If we run into a problem like that, we'll have to look at it :D But for now, we care about the 100th term, which is the second to last term.

    • 2 years ago
  14. zepdrix Group Title
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    If you look at pascal's triangle, what can you say about the second AND second to last numbers in any given row?

    • 2 years ago
  15. lilsis76 Group Title
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    well the second number would be the 1, and the second to last number would be i think the 100-1?

    • 2 years ago
  16. zepdrix Group Title
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    Mmm it might be a little confusing because we're actually counting from ZERO, not from 1. So the first row in pascal's triangle corresponds to a binomial of degree zero. So look at the forth row, which corresponds to a binomial of degree 3. The terms are 1 3 3 1 The next row would produce, 1 4 6 4 1. The second and second to last number are 4 yes? :D The 101st row (Which corresponds to a 100th degree binomial) has terms, 1 100 ... 100 1 So the second and second to last numbers are 100 right? :D

    • 2 years ago
  17. lilsis76 Group Title
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    OH!, uh ya i think so cuz, the first set will always be one, and the next number would be the number we are looking for right?

    • 2 years ago
  18. zepdrix Group Title
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    Yes, the coefficient at least :) The coefficient on the second and second to last term will always match the DEGREE of the binomial.

    • 2 years ago
  19. zepdrix Group Title
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    So hmmm.. the 3rd degree binomial (4th row in pascal's triangle) had 4 terms in it. The next row will have 5 terms in it. The 100th degree binomial (101st row in pascal's triangle) will have 101 terms in it. We want the 100th term. So we want the 2nd to last term in that sequence.

    • 2 years ago
  20. zepdrix Group Title
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    These are the last 2 terms in the sequence.. \[\large (1+x)^{100}=.....+100(1^1)x^{99}+1(1^0)x^{100}\] Understand where they're coming from? I know this can be a bit confusing :c

    • 2 years ago
  21. zepdrix Group Title
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    More formally you can generalize it like this, \[\huge (1+x)^n=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)1^{n-k}x^k\] Where \[\large \left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\frac{ n! }{ k!(n-k)! }\right)\]

    • 2 years ago
  22. zepdrix Group Title
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    So the 100th term in the sequence would be when n=100, and k=99 (because k starts counting from 0, not from 1). \[\large \frac{ 100! }{ 99!(100-99)! }1^{100-99}x^{99}=\frac{100*99!}{99!}1^1x^{99}\] \[\huge =100x^{99}\]

    • 2 years ago
  23. zepdrix Group Title
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    It's quite a bit to chew on :\ maybe take a look it and try to get through it! :D There is the answer at least. :)

    • 2 years ago
  24. lilsis76 Group Title
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    sorry im back. connection was weird

    • 2 years ago
  25. lilsis76 Group Title
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    okay thank you. an this is super confusing im trying to figure it out still by looking at what you put

    • 2 years ago
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