1⼀2+(1⼀2)^2+(1⼀2)^3+(1⼀2)^4+…+(1⼀2)^=

跪求,今天2点之前要。加分
2024-12-17 17:12:17
推荐回答(3个)
回答1:

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

回答2:

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

回答3:

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.