perplexus.info :: Just Math : Binomial Baffle

Binomial Baffle (Posted on 2014-12-30)

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.

See The Solution Submitted by K Sengupta

Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

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

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info