已知数列{an}满足Sn=1+1?
已知数列{an}满足Sn=1+1/4an,则an=
令n=1,得a1=1+(1/4)a1,(3/4)a1=1,a1=4/3. n>1时S<n-1>=1+(1/4)a<n-1>, 与原式相减得an=(1/4)an-(1/4)a<n-1>, (3/4)an=-(1/4)a<n-1>, an=(-1/3)a<n-1> =(-1/3)^(n-1)×a1 =(4/3)×(-1/3)^(n-1).
答:解: ∵Sn=2n-an ∴S1=a1=2-a1 a1=1 Sn=2n-an S(n-1)=2(n-1)-a(n-1) Sn-S(n-1)=an=a(n-1)-...详情>>
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