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Binomial theorum

  1. Jun 16, 2006 #1
    i an havign trouble solving this qs

    if nC0 + nC1 + nC2 +...+ nCn = 256 find the value of n

    all help appreciated:smile:
     
  2. jcsd
  3. Jun 16, 2006 #2

    StatusX

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    You know the binomial theorem, so what values of x and y have the expansion of (x+y)^n equal to that sum?
     
  4. Jun 16, 2006 #3
    i don't get it, i know that teh values of x and y shoudl be one, but if it is equal to 256 how can i put that ?
     
  5. Jun 16, 2006 #4

    StatusX

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    Right, so what does n have to be for (1+1)^n to equal 256?
     
  6. Jun 17, 2006 #5

    HallsofIvy

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    Do you understand that the crucial question here is "What is 1+ 1?":rofl:
     
  7. Jun 18, 2006 #6
    its a series question . i forgot how to do it..sorry.
     
  8. Jun 18, 2006 #7
    cool i solved it lol. took less than 5 mins tried and error on calc.. there is a proper way to solve it... well n=8 i work it out by put numbers into n@_@ yeh 8 is correct value.


    omfg im sorry guys it oready been solve... by (1+1)^n=256 <--- how did dat work out@_@ well i did tried my best@_@
     
    Last edited: Jun 18, 2006
  9. Jun 18, 2006 #8

    HallsofIvy

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    Excuse me? 'Trial and error'? The whole question was "for what n does n does 2n= 256. How long does that take to calculate?
    22= 4, 23= 8, 24= 16, 25= 32, 26= 64, 27= 128, 28= 256. Well, gosh, I guess n= 8 so that 2n=256!
     
  10. Jun 18, 2006 #9
    thnx...i get it now
     
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