Question

求数列的和：Sn=1+(1+1/2)+(1+1/2+1/4)+…+(1+1/2+1/4+…+1/2^n-1)

求详细过程 求Sn

Answer 1

an=1+1/2+1/4+...+1/2^(n-1)=[1-(1/2)^n]/(1-1/2)=(1/2)[1-(1/2)^n]
Sn=a1+a2+...+an
=(1/2)[1-(1/2)^1]+(1/2)[1-(1/2)^2]+...+(1/2)[1-(1/2)^n]
=n/2-(1/2^2+1/2^3+...+1/2^(n+1))
=n/2-(1/4)[1-(1/2)^n]/(1-1/2)
=n/2-(1/2)[1-(1/2)^n]
=n/2+1/2^(n+1)-1/2