Question

记a为(1-x)^n展开式中x^2的系数,且lim[a/(bn^2+1)]=1则实数b=________

Answer 1

T(r+1)=C(n,r)*1^(n-r)*(-x)^r.

由题设知：r=2.

故,a=C(n,2)=n(n-1)/2=(n^2-n)/2.

lim[a/(bn^2+1)=1.

即,lim{[ (n^2-n)/2] /(bn^2+1)}=1.

lim(1-1/n)/(2b+1/n^2)=1.

当n--->∞时,得：1/2b=1.

故,b=1/2.