(40) A.maths (Binomial Theorem) 1Qs
2007-10-30 6:38 am
Q25 only.
If you can't see the pict. clearly, you can go to http://x4b.xanga.com/df3c53e477734154699629/m115723705.jpg
Thanks for your help...
回答 (2)
2007-10-30 6:48 am
✔ 最佳答案
The image is indeed very difficult to see...n (n C 2) = (n - 1)[n + (3/2)]
n { n!/ [(n - 2)! 2!] } = (n - 1)[n + (3/2)]
n (n) (n - 1) = 2 (n - 1) [n + (3/2)]
For n =/= 1,
n^2 = 2 [n + (3/2)]
n^2 - 2n - 3 = 0
(n + 1) (n - 3) = 0
n = -1 (rejected as n is a positive integer.) or n = 3
Therefore, n = 3.
If there is a mistake, please inform me.
參考: My Maths knowledge
2007-10-30 6:45 am
n Cn2 = (n–1)(n + 3/2)(n≧2)
n[n(n–1)/2] = (n–1)(n + 3/2)
n2(n - 1)/2 = (n–1)(n + 3/2)(∵n≠1)
n2/2 = n + 3/2
2n2 = n + 3
2n2–n–3 = 0
(n–3)(n + 1) = 0
n = 3 or -1 ( but which is rejected)
∴n = 3
n[n(n–1)/2] = (n–1)(n + 3/2)
n2(n - 1)/2 = (n–1)(n + 3/2)(∵n≠1)
n2/2 = n + 3/2
2n2 = n + 3
2n2–n–3 = 0
(n–3)(n + 1) = 0
n = 3 or -1 ( but which is rejected)
∴n = 3
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