Question

一道关于高中数学的等比数列的题数列{a的第n项}的前n项和计为Sn,已知a1=1,a的第(n+1)项=Sn*(n+2)/

Answer 1

因为A(n+1) = (n+2)/n * Sn

所以Sn = n*A(n+1) / (n+2)

S(n-1) = (n-1)*An / (n+1)

所以An = Sn - S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1) * An

所以2n/(n+1) * An = n/(n+2) * A(n+1)

即A(n+1)/An = (2n+4)/(n+1)

所以(Sn/n) / (S(n-1)/(n-1)) = ( A(n+1)/(n+2) ) / ( An / (n+1))

= A(n+1)/An * (n+1)/(n+2)

= (2n+4)/(n+1) * (n+1)/(n+2) = 2

所以Sn/n是以2为公比的等比数列

(2)

因为Sn/n是以2为公比的等比数列,首项为S1/1=S1=A1=1

所以Sn/n的通项公式是2^(n-1)

所以Sn = n*2^(n-1)

S(n-1) = (n-1)*2^(n-2)

所以An = Sn - S(n-1) = n*2^(n-1) - (n-1)*2^(n-2)

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

= n*2^(n-2) + 2^(n-2)

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

当n=1时也满足,所以通项公式为An = (n+1) * 2^(n-2)