已知数列{an}满足a1=2,an=1/2an+1-2^n(n∈N+)求前n项和Sn

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$已知数列{an}满足a1=2,an=1/2an+1-2^n(n∈N+)求前n项和Sn$

已知数列{an}满足a1=2,an=1/2an+1-2^n(n∈N+)求前n项和Sn
已知数列{an}满足a1=2,an=1/2an+1-2^n(n∈N+)求前n项和Sn

已知数列{an}满足a1=2,an=1/2an+1-2^n(n∈N+)求前n项和Sn
由已知an=1/2a(n+1)-2^n
两边同时除以2^n得到,
a(n+1)/2^(n+1)=an/2^n+1
可以看出an/2^n是公差为1的等差数列,
令bn=an/2^n,那么b1=a1/2=1,
bn=n
an/2^n=bn=n,
an=n*2^n,
所以sn=a1+a2+...+an=1*2+2*2^2+3*2^3+.n*2^n
2sn=1*2^2+2*2^3+3*2^4+.n*2^(n+1)
两式相减得到,
sn=n*2^(n+1)-(2+2^2+2^3+.+2^n)=(n-1)*2^(n+1)-2

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