Question

二项式定理题 急已知an=1+q+q^2+……+q^(n-1)（n属于正整数,q不等于正负1）,An=nC1*a1+nC

Answer 1

an=(1-q^n)/(1-q)

An=nC1*a1+nC2*a2+……+nCn*an

=[nC0*(1-q^0)+nC1*(1-q)+nC2*(1-q^2)+……+nCn*(1-q^n)]/(1-q)

={-[nC0*(q^0)+nC1*(q)+nC2*(q^2)+……+nCn*(q^n)]+[nC0+nC1+nC2+……+nCn]}/(1-q)

=[-(1+q)^n+2^n]/(1-q)

bn=A(n)/2^n-A(n-1)/2^(n-1)

=[1-((1+q)/2)^n]/(1-q)-[1-((1+q)/2)^(n-1)]/(1-q)

=[((1+q)/2)^(n-1)-((1+q)/2)^n]/(1-q)

=((1+q)/2)^(n-1)*(1-(1+q)/2)/(1-q)

=((1+q)/2)^(n-1)/2

为等比数列