设f(x)=(1+x)+(1+x)^2+...+(1+x)^n,f(x)中x^2的系数为Tn,则limTn/(n^3+2n)等于:A:1/3 B:1/6C:1 D:2
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![设f(x)=(1+x)+(1+x)^2+...+(1+x)^n,f(x)中x^2的系数为Tn,则limTn/(n^3+2n)等于:A:1/3 B:1/6C:1 D:2](/uploads/image/z/5251034-2-4.jpg?t=%E8%AE%BEf%28x%29%3D%281%2Bx%29%2B%281%2Bx%29%5E2%2B...%2B%281%2Bx%29%5En%2Cf%28x%29%E4%B8%ADx%5E2%E7%9A%84%E7%B3%BB%E6%95%B0%E4%B8%BATn%2C%E5%88%99limTn%2F%28n%5E3%2B2n%29%E7%AD%89%E4%BA%8E%EF%BC%9AA%3A1%2F3+B%3A1%2F6C%3A1+D%3A2)
设f(x)=(1+x)+(1+x)^2+...+(1+x)^n,f(x)中x^2的系数为Tn,则limTn/(n^3+2n)等于:A:1/3 B:1/6C:1 D:2
设f(x)=(1+x)+(1+x)^2+...+(1+x)^n,f(x)中x^2的系数为Tn,则limTn/(n^3+2n)等于:
A:1/3 B:1/6
C:1 D:2
设f(x)=(1+x)+(1+x)^2+...+(1+x)^n,f(x)中x^2的系数为Tn,则limTn/(n^3+2n)等于:A:1/3 B:1/6C:1 D:2
根据二项式定理,
(1+x)^n中x^2的系数为C(n,2)=n (n≥2)
故f(x)中x^2的系数为
Tn=C(2,2)+C(3,2)+...+C(n,2)
=2*1/2+3*2/2+...+n(n-1)/2
=(2*1+3*2+...+n(n-1))/2
=[2^2-2+3^2-3+...+n^2-n]/2
=[2^2+3^2+...+n^2-(2+3+...+n)]/2
=[1^2+2^2+3^2+...+n^2-(1+2+3+...+n)]/2
=[n(n+1)(2n+1)/6-n(n+1)/2]/2
容易看出其中n^3的系数为2/6/2=1/6,
因此
lim(n->∞) Tn/(n^3+2n)
=(1/6) / 1
=1/6
选B!