数列an满足条件a1=1,an=a(n-1)+(1/3)^(n-1) (n=2,3,…)
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解:(1)∵数列an满足条件a1=1,an=a(n-1)+(1/3)^(n-1) (n=2,3,…)
∴a2=a1+1/3=1+1/3
a3=a2+(1/3)^2=1+1/3+(1/3)^2
a4=a3+(1/3)^3=1+1/3+(1/3)^2+(1/3)^3
........
a(n-1)=a(n-2)+(1/3)^(n-2)=1+1/3+(1/3)^2+(1/3)^3+....+(1/3)^(n-2)
故an=a(n-1)+(1/3)^(n-1)
=1+1/3+(1/3)^2+(1/3)^3+....+(1/3)^(n-2)+(1/3)^(n-1)
=[1-(1/3)^n]/(1-1/3)
=3[1-(1/3)^n]/2
(2)a1+a2+a3+…+an
=3(1-1/3)/2+3[1-(1/3)^2]/2+3[1-(1/3)^3]/2+....+3[1-(1/3)^n]/2
=3n/2-3[1/3+(1/3)^2+(1/3)^3+....+(1/3)^n]/2
=3n/2-[1+1/3+(1/3)^2+....+(1/3)^(n-1)]/2
=3n/2-[1-(1/3)^n]/[2(1-1/3)]
=3n/2-3[1-(1/3)^n]/4
=[6n-3+(1/3)^n]/4
∴a2=a1+1/3=1+1/3
a3=a2+(1/3)^2=1+1/3+(1/3)^2
a4=a3+(1/3)^3=1+1/3+(1/3)^2+(1/3)^3
........
a(n-1)=a(n-2)+(1/3)^(n-2)=1+1/3+(1/3)^2+(1/3)^3+....+(1/3)^(n-2)
故an=a(n-1)+(1/3)^(n-1)
=1+1/3+(1/3)^2+(1/3)^3+....+(1/3)^(n-2)+(1/3)^(n-1)
=[1-(1/3)^n]/(1-1/3)
=3[1-(1/3)^n]/2
(2)a1+a2+a3+…+an
=3(1-1/3)/2+3[1-(1/3)^2]/2+3[1-(1/3)^3]/2+....+3[1-(1/3)^n]/2
=3n/2-3[1/3+(1/3)^2+(1/3)^3+....+(1/3)^n]/2
=3n/2-[1+1/3+(1/3)^2+....+(1/3)^(n-1)]/2
=3n/2-[1-(1/3)^n]/[2(1-1/3)]
=3n/2-3[1-(1/3)^n]/4
=[6n-3+(1/3)^n]/4
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