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lilsis76

  • 2 years ago

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

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  1. surdawi
    • 2 years ago
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    is x the number of terms?

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

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

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

  5. zepdrix
    • 2 years ago
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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.

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

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

  8. zepdrix
    • 2 years ago
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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

  9. zepdrix
    • 2 years ago
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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.

  10. zepdrix
    • 2 years ago
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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

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

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

  13. zepdrix
    • 2 years ago
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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.

  14. zepdrix
    • 2 years ago
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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?

  15. lilsis76
    • 2 years ago
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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?

  16. zepdrix
    • 2 years ago
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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

  17. lilsis76
    • 2 years ago
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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?

  18. zepdrix
    • 2 years ago
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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.

  19. zepdrix
    • 2 years ago
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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.

  20. zepdrix
    • 2 years ago
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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

  21. zepdrix
    • 2 years ago
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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 \[\large \left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\frac{ n! }{ k!(n-k)! }\right)\]

  22. zepdrix
    • 2 years ago
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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). \[\large \frac{ 100! }{ 99!(100-99)! }1^{100-99}x^{99}=\frac{100*99!}{99!}1^1x^{99}\] \[\huge =100x^{99}\]

  23. zepdrix
    • 2 years ago
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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. :)

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

  25. lilsis76
    • 2 years ago
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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

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