About binomial theorem

偉聰

2006-12-06 9:46 am

1. In the expansion of(1+x)^n , two times the coefficient of x^5 is equal to the sum of the coefficients of x^4andx^6 . Find the possible values of n.
2. In the expansion of (x^2+2)^n in descending powers of x, where n is a positive integer, the coefficient of the third term is 40. Find the value of n and the coefficient of x^4.

THx

回答 (1)

myisland8132

2006-12-06 11:19 am

✔ 最佳答案

1
the coefficient of x^4=nC4
the coefficient of x^5=nC5
the coefficient of x^6=nC6
2nC5=nC4+nC6
2/5!(n-5)!=1/4!(n-4)!+1/6!(n-6)!
1/[60(n-5)!]=1/[24(n-4)!]+1/[720(n-6)!]
1/[60(n-5)]=1/[24(n-4)(n-5)]+1/720
12/(n-5)=30/[(n-4)(n-5)]+1
12(n-4)=30+(n-4)(n-5)
12n-48=30+n^2-9n+20
n^2-21n+98=0
(n-12)(n-9)=0
n=9 or 12
2
(x^2+2)^n
=x^2n+2n(x^2(n-1))+4(nC2)(x^2(n-2))+...
so
4(nC2)=40
nC2=10
n=5
(x^2+2)^5
=x^10+20(x^8)+40(x^6)+8(10)(x^4)+...
the coefficient of x^4
=80

2006-12-06 03:22:13 補充：
第一條應是n^2-21n 98=0(n-14)(n-7)=0n=7 or 14

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