解:设S=1/2+(1/2)^2+(1/2)^3+(1/2)^4+…+(1/2)^n·············1
则2S=1+(1/2)^2+(1/2)^3+(1/2)^4+…+(1/2)^(n-1)··············2
2式减1式得:
S=1/2-(1/2)^n
等比数列的求和公式为Sn=a1(1-q^n)/(1-q),本题的a1=q=1/2,所以原式=1-(1/2)^n
典型的等比数列....前n项和的公式是.Sn=a1(1-q^n)/(1-q) a1是首项即1/2.. q是公比1/2.