Binomial Expansion(9)

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2010-03-07 6:23 am

(a) Expand (1-x+nx^2)^n in ascending powers of x as far as the term in x^3.
(b) Given that the coefficient of x^2 in the expansion of (1-2x)(1+2x+nx^2)^n is 45. Find the value of n and the coefficient of x^3.

回答 (1)

So Much

2010-03-07 9:31 am

✔ 最佳答案

(1-x+nx^2)^n

=(1+x(nx-1))^n

=(nC0)+(nC1)x(nx-1)+(nC2)x^2(nx-1)^2+(nC3)(x^3)(nx-1)^3+...

=1+n(nx^2-x)+(1/2)n(n-1)(x^2)(1-2nx+...)+(1/3)n(n-1)(n-2)x^3(-1)+...

=1+n^2x^2-nx+(1/2)n(n-1)x^2-n^2(n-1)x^3-(1/3)n(n-1)(n-2)x^3+...

=1-nx+((1/2)(n^2-n)+n^2)x^2+(-(1/3)n(n-1)(n-2)+n^2(n-1))x^3+...

=1-nx+(1/2)(3n^2-n)x^2+(1/6)(n)(7n-2)(1-n)x^3+...//

2010-03-07 01:34:35 補充：
(1-2x)(1+2x+nx^2)^n

=(1-2x)(1-nx+(1/2)(3n^2-n)x^2+(1/6)(n)(7n-2)(1-n)x^3+...)

=...+((1/2)(3n^2-n)+2n)x^2+...

then (1/2)(3n^2-n+4n)=45

=>n^2+n=30

=>n^2+n-30=0

=>(n+6)(n-5)=0

=>n=-6 ( rejected) or n=5//

2010-03-07 01:35:19 補充：
coef or x^3 term is -180//

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