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Binomial Baffle (Posted on 2014-12-30) Difficulty: 3 of 5
If we expand (1 + 0.2)2014 by the binomial theorem, we get:
A(0) + A(1) + A(2) + .......+ A(2014) where:
A(j) = C(2014, j) *(0.2)j and:
C(x, y) = x!/(y!*(x-y)!)

Find the value of j for which A(j) is maximum.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Paper and pencil solution (spoiler) Comment 2 of 2 |
We are looking for the largest j for which A(j)/A(j-1) is > 1.

Well, A(j)/A(j-1)
  = .2*(2014!/j!(2014-j)!)/((2014!/(j-1)!(2014-j+1)!)
  = .2(2015-j)/j
  
Setting this equal to 1 and solving gives j = 2015/6 = 335.83

So our maximum value is either at 335 or 336.

A(336)/A(335) = .2*(2015-336)/336 = 0.9994
A(335)/A(334) = .2*(2015-335)/335 = 1.0030

So A increases until j = 335, and decreases thereafter.
Maximum is at j = 335

  Posted by Steve Herman on 2014-12-30 18:40:07
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