等比数列{an}满足:a1=1/2,且an-an-1=1/2的n次方,求an

an-a[n-1]=1/2^n
a[n-1]-a[n-2]=1/2^(n-1)
...
a2-a1=1/2^2
以上各式相加得:
an-a1=(1/2^2+...+1/2^n)=1/2^2*(1-1/2^(n-1))/(1-1/2)=1/2*(1-2/2^n)=1/2-1/2^n
所以,an=1/2-1/2^n+a1=1-1/2^n.