Question

数列中,a1=1,且a1,a2-a1,a3-a2,…an-an-1,…是公比为1/3的等比数列,那么an=?

Answer 1

(a2-a1)/a1=1/3

an-an-1=[1-(1/3)^n]/(1-1/3)

=[1-(1/3)^n]/(2/3)

=3*[1-(1/3)^n]/2

=[3-(1/3)^(n-1)]/2

=3/2-(1/3)^(n-1)/2

an-an-1=3/2-(1/3)^(n-1)/2

a(n-1)-a(n-2)=3/2-(1/3)^(n-2)/2

.

a3-a2=3/2-(1/3)^2/2

a2-a1=3/2-(1/3)^1/2=3/2-1/6

以上等式相加得

an-a1=3/2-(1/3)^(n-2)/2+3/2-(1/3)^(n-2)/2

+.3/2-(1/3)^1/2+3/2-(1/3)^1/2

=3n/2+1/6*[1-(1/3)^(n-1)]/(1-1/3)

=3n/2+1/2*[1-(1/3)^n]/(2/3)

=3n/2+3*[1-(1/3)^n]/4

=3n/2+3/4-(1/3)^(n-1)/4

an=3n/2+3/4-(1/3)^(n-1)/4+1

an=3n/2+7/4-(1/3)^(n-1)/4