设数列{an}满足a1=a,a(n+1)=can+1-c,n∈N+,其中a,c为实数,且c≠0.(1)设a=0.5,c=0.5,bn=n(1-an),n∈N+成立,求数列{bn}的前n项和Sn;(2)若0

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设数列{an}满足a1=a,a(n+1)=can+1-c,n∈N+,其中a,c为实数,且c≠0.(1)设a=0.5,c=0.5,bn=n(1-an),n∈N+成立,求数列{bn}的前n项和Sn;(2)若0
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设数列{an}满足a1=a,a(n+1)=can+1-c,n∈N+,其中a,c为实数,且c≠0.(1)设a=0.5,c=0.5,bn=n(1-an),n∈N+成立,求数列{bn}的前n项和Sn;(2)若0
设数列{an}满足a1=a,a(n+1)=can+1-c,n∈N+,其中a,c为实数,且c≠0.
(1)设a=0.5,c=0.5,bn=n(1-an),n∈N+成立,求数列{bn}的前n项和Sn;
(2)若0

设数列{an}满足a1=a,a(n+1)=can+1-c,n∈N+,其中a,c为实数,且c≠0.(1)设a=0.5,c=0.5,bn=n(1-an),n∈N+成立,求数列{bn}的前n项和Sn;(2)若0
a=c=1/2
a(n+1)=an/2+1/2
1-a(n+1)=(1-an)/2
1-a1=1-a=1/2
1-an=1/(2^n)
bn=n/(2^n)
由错位相减法
Sn=2-(n+2)/(2^n)
(2)
0

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