计算（(1/3)^2）×4^0+（(1/3)^4）×4^1+（(1/3)^6）×4^2+…+（(1/3)^2n）×4^(

Sn=a1(1-q^n)/(1-q)=(a1-an×q)/(1-q) a1=[(1/3)^2]×4^0=1/9q=[(1/3)^2]×4=4/9an=[(1/3)^2n]×4^(n-1)=[4^(n-1)]/(9^n)an×q=[4^(n-1)]/(9^n)×4/9=(4^n)/[9^(n+1)]1-q=1-4/9=5/9Sn=(a1-an×q)/(1-q)={1/9-(4^n)/[9^(n+1)]}/(5/9)=(9^n-4^n)/(5×9^n)=1/5×[1-(4/9)^n]=1/5-(4^n)/[5×(9^n)] 小蛤