In the expansion of (1+x)^n, the sum of the coefficients of x^4 and x^6 is twice the coefficients of x^5. If n is an integer and n ＞ 5, find the values of n.
* Please state the steps in detail
F.4 --- binomial theorem
2009-10-26 9:32 am
回答 (1)
2009-10-26 9:40 am
✔ 最佳答案
Coefficient of x4 = nC4Coefficient of x5 = nC5
Coefficient of x6 = nC6
Hence:
nC4 + nC6 = 2 x nC5
n!/[4! (n - 4)!] + n!/[6! (n - 6)!] = 2 (n!)/[5! (n - 5)!]
1/[4! (n - 4)!] + 1/[6! (n - 6)!] = 2/[5! (n - 5)!]
Multiplying both sides by (n - 4)! x 6!:
6 x 5 + (n - 4)(n - 5) = 2 x 6 x (n - 4)
30 + n2 - 9n + 20 = 12n - 48
n2 - 21n + 98 = 0
(n - 14)(n - 7) = 0
n = 7 or 14
參考: Myself
收錄日期: 2021-04-13 16:54:18
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