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Old 08-01-2007   #1
Zaid h
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Default Binomial theorm (Statistics, Data management) Question.?

Determine the middle term in the expansion of (x-y^3)^16 (note: think about which term is the middle one)Thanks to anyone that can help me, appreciate it
Old 08-01-2007   #2
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C(n,i)=n!/(n-i)!i!C(16,i)=16!/(16-i)!i!(x-y^3)^16=(-1)^i C(16,i) x^(16-i)(y^3)^iThenC(16,0) x^16(y^3)^0-C(16,1) x^15(y^3)^1+C(16,2)x^14(y^3)^2-C(16,3) x^13(y^3)^3...(-1)^i C(16,i) x^(16-i)(y^3)^i
Old 08-01-2007   #3
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In (x+y)^n there are (n+1) terms Therefore ^ 16 means that there are 17 terms & hence 9th term is the mid termThe 9th term is 16 C8 * (x)^(16-8).(-y^3)^8= 16C8 { x^8.y^24)}=(16C8)x^8.y^24where 16C8= (16*15*14*13*12*11*10*9)/(1*2*3*4*5*6*7*9*10)

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