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anonymous

  • 4 years ago

the coefficient of x^2 in the expansion of (1 + x/5)^n, where n is a positive integer, is 3/5. Find the value of n. How to do this???? :P

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  1. anonymous
    • 4 years ago
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    \[(1+x)^n=1+nx+\dbinom{n}{2}x^2+...\] in your case you will have \[1+n\frac{x}{5}+\dbinom{n}{2}\frac{x^2}{25}+ ...\]

  2. anonymous
    • 4 years ago
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    so basically i need to multiply 1/25 and \[\left(\begin{matrix}n \\ 2\end{matrix}\right)\], but how do I do that?

  3. anonymous
    • 4 years ago
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    and equate it to 3/5.

  4. anonymous
    • 4 years ago
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    so you know that \[\frac{n(n-1)}{2}\times \frac{1}{25}=\frac{3}{5}\] \[\frac{n(n-1)}{2\times 25}=\frac{3}{5}\] \[\frac{n(n-1)}{10}=3\] \[n(n-1)=30\]

  5. anonymous
    • 4 years ago
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    and since n is a whole number, only possible answer is \[n=6\]

  6. anonymous
    • 4 years ago
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    wait, sorry, I don't get why n(n - 1)/2

  7. anonymous
    • 4 years ago
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    oh ok lets go slower

  8. anonymous
    • 4 years ago
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    to compute \[\dbinom{n}{k}\] you do not really use \[\frac{n!}{k!(n-k)!}\] because it it is too much work, you just cancel. but we can write it out anyway

  9. anonymous
    • 4 years ago
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    \[\dbinom{n}{2}=\frac{n!}{2!(n-2)!}\] ok?

  10. anonymous
    • 4 years ago
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    ohhh. so if it was the coefficient of x^3, then we would do n(n - 1)(n - 3)/3 ??

  11. anonymous
    • 4 years ago
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    then numerator is \[n\times (n-1)\times (n-2)!\] so you can cancel the \[(n-2)!\] and get \[\frac{n(n-1)}{2}\]

  12. anonymous
    • 4 years ago
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    it would be \[\frac{n(n-1)(n-2)}{3!}\]

  13. anonymous
    • 4 years ago
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    don't forget the factorial in the denominator

  14. anonymous
    • 4 years ago
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    ^sorry, that's what I meant to type :P Got it.

  15. anonymous
    • 4 years ago
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    great!

  16. anonymous
    • 4 years ago
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    So, what's up, satellite??? haven't been here in a while :P

  17. anonymous
    • 4 years ago
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    not much. where you been?

  18. anonymous
    • 4 years ago
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    Sitting at home, studying for my mocks -_-. Ohhh, and i've got braces. i look WEIRD.

  19. anonymous
    • 4 years ago
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    been a month it seems

  20. anonymous
    • 4 years ago
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    ouch. did they hurt?

  21. anonymous
    • 4 years ago
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    well, i know. ^nopee, not much, but after a few hours they're tickling me :P And I feel like a kid, 'cause over here most people had their braces off by the 7th / 9th grade, and i'm starting in the 10th.

  22. anonymous
    • 4 years ago
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    But i've got a pretty blue, AND it's only for a year, so it's cool.

  23. anonymous
    • 4 years ago
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    well at least they are a color and not those awful railroad tracks!

  24. anonymous
    • 4 years ago
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    *7th/8th grade. I KNOW, those are horrid. My brother didn't put any colours - I killed him for it. Idiot.

  25. anonymous
    • 4 years ago
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    More than that, I know some people who have gotton the clear, translucent braces, not the metal ones, and they've put colours on them. Kinda kills the whole point of see-through ones, doesn't it???

  26. anonymous
    • 4 years ago
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    the new thing seems to be to make you have them twice. once around 6th /7th, then a year off, then back again. i think it is just a way to get more money out of parents

  27. anonymous
    • 4 years ago
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    What cheapsters.

  28. anonymous
    • 4 years ago
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    See??/ i'm such a nice girl, I became your fan :P

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