Let Br denote the coefficient of x^r in the expansion of (1+x)^24 and B(r+2):Br=57:7. find the value of r.
P.S. B(r+2) 同 Br 既 r不是乘...
ans=5
plz show the step
(1+x)^24 FIND
2008-06-20 12:09 am
回答 (2)
2008-06-20 12:27 am
✔ 最佳答案
Br = 24Cr = 24!/(r!(24-r)!) ,B(r+2) = 24C(r+2) = 24!/((r+2)!(24-r-2)!)
B(r+2)/Br
= r!(24-r)!/((r+2)!(24-r-2)!)
= (24-r)(24-r-1)/((r+1)(r+2))
= 57/7
Therefore,
57(r+1)(r+2) = 7(24-r)(24-r-1) = 7(24-r)(23-r)
57r^2 + 171r + 114 = 3864 -329r + 7r^2
50r^2 + 500r - 3750 = 0
r^2 + 10r -75 = 0
(r+15)(r-5)=0
r = -15 or r = 5
Since r cannot be negative, r = 5
2008-06-20 2:04 am
under Pascal's triangle
we know expanding (1+x)^24
we will get something like:
24C0*x^0 + 24C1*x^1 + 24C2*x^2 + ... + 24C24*x^24
已知:
B(r+2)/Br=57/7
so
24C(r+2)/24Cr = 57/7
{24!/[(r+2)!*(24-r-2)!]}/{24!/[r!*(24-r)!]} = 57/7
[r!*(24-r)!]/[(r+2)!*(22-r)!] = 57/7
r!*(24-r)! / {[(r+2)(r+1)r!]*(24-r)!]/(24-r)(23-r)} = 57/7
[(24-r)(23-r)]/[(r+1)(r+2)] = 57/7
7(24-r)(23-r) = 57(r+1)(r+2)
7r^2 - 329r + 3864 = 57r^2 + 171r +114
50r^2 + 500r - 3750 = 0
r=5 or r=-15
we know expanding (1+x)^24
we will get something like:
24C0*x^0 + 24C1*x^1 + 24C2*x^2 + ... + 24C24*x^24
已知:
B(r+2)/Br=57/7
so
24C(r+2)/24Cr = 57/7
{24!/[(r+2)!*(24-r-2)!]}/{24!/[r!*(24-r)!]} = 57/7
[r!*(24-r)!]/[(r+2)!*(22-r)!] = 57/7
r!*(24-r)! / {[(r+2)(r+1)r!]*(24-r)!]/(24-r)(23-r)} = 57/7
[(24-r)(23-r)]/[(r+1)(r+2)] = 57/7
7(24-r)(23-r) = 57(r+1)(r+2)
7r^2 - 329r + 3864 = 57r^2 + 171r +114
50r^2 + 500r - 3750 = 0
r=5 or r=-15
收錄日期: 2021-04-16 22:52:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080619000051KK01511