Question

设数列{an}的前n项和为Sn，且满足Sn=2-an，n=1，2，3。。。。。 求数列an通

设数列{an}的前n项和为Sn，且满足Sn=2-an，n=1，2，3。。。。。 求数列an通项公式；若数列bn满足b1=1，且b(n+1)=bn+an，求bn通项公式；设cn=n(3-bn)，求数列cn前n项和Tn

Answer 1

Sn=2-an (1)

put n=1
a1=1
S(n-1)=2-a(n-1) (2)

(1) -(2)
an = -an+a(n-1)
an/a(n-1) = 1/2
an/a1 = (1/2)^(n-1)
an = (1/2)^(n-1)
b(n+1) = bn +an
b(n+1)-bn = (1/2)^(n-1)
bn -b(n-1) = (1/2)^(n-2)
bn-b1 = 1+(1/2)+..+(1/2)^(n-2)
= 2(1-(1/2)^(n-1))
bn = 1+ 2(1-(1/2)^(n-1))
= 3 - (1/2)^(n-2)
cn= n(3-bn)
= n(1/2)^(n-2)
=2(n.(1/2)^(n-1))
consider
1+x+x^2+.+x^n = (x^(n+1) -1)/(x-1)
1+2x+..+nx^(n-1) = [(x^(n+1) -1)/(x-1)]'
= [nx^(n+1) -(n+1)x^n +1]/(x-1)^2
put x= 1/2
1.(1/2)^0+ 2(1/2)+..+n(1/2)^(n-1)
=4[n(1/2)^(n+1) -(n+1)(1/2)^n +1]
=4- (2n+4)(1/2)^n
Tn = c1+c2+..+cn
=2[4- (2n+4)(1/2)^n]
= 8 -(n+2)(1/2)^(n-2)