数列{an}中满足a(n+1)=an+(1/2)^n,a1=1求an和Sn

显示全部楼层 · 2009-2-26 21:42:34

a(n+1)-an=(1/2)^(n-1)所以an-a(n-1)=(1/2)^(n-1)a(n-1)-a(n-2)=(1/)^(n-2)……a2-a1=(1/2)^(1-1)相加an-a1=(1/2)^0+……+(1/2)^(n-1)后面有n项，所以an-a1=(1/2)^0*[1-(1/2)^n]/(1-1/2)=2-2*(1/2)^na1=1所以an=3-2*(1/2)^nSn=3*n-2*[1/2+……+(1/2)^n]=3n-2*(1/2)*[1-(1/2)^n]/(1-1/2)=3n-2+2*(1/2)^n...

千问 · 2009-2-26 21:42:34

a(n+1)-an=(1/2)^n;an-a(n-1)=(1/2)^(n-1);............a2-a1=(1/2)^1;a1=1;相加有a(n+1)=（1/2）^0+（1/2)^1+....(1/2)^(n+1)=2-(1/2)^n;an=2-(1/2)^(n-1);(n>=1)a1=1;Sn=(2-1)+(2-1...