已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N)设数列｛an｝的前n项和为SN,是否存在实数M,使得对于任意的n属于N,恒有SN

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已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列｛an｝的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)
设数列｛an｝的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN

已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列｛an｝的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
a(n+1)=1/2 (2n+3)/(2n+1) an
=1/2*1/2 *(2n+3)/(2n+1) *(2n+1)/(2n-1) *a(n-1)
=.
=(1/2)^n *(2n+3)/(2n+1) *(2n+1)/(2n-1) *.* 5/3 *a1
=(1/2)^n *(2n+3)/3 *3/2
=(2n+3) *(1/2)^(n+1)
an=(2n+1)*(1/2)^n
Sn=3*1/2 +5*(1/2)^2+7*(1/2)^3+.+(2n+1)*(1/2)^n
1/2 Sn=3*(1/2)^2+5*(1/2)^3+.+(2n-1)*(1/2)^n+(2n+1)*(1/2)^(n+1)
1/2Sn=Sn-1/2Sn=3*1/2 -2((1/2)^2+(1/2)^3+.+(1/2)^n) -(2n+1)*(1/2)^(n+1)
=3/2-(2n+1)*(1/2)^(n+1) +1/2 *(1-(1/2)^(n-1)) /(1-1/2)
=3/2+1-(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
=5/2 -(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
所以Sn=5-2(1/2)^(n-1)-2(2n+1)*(1/2)^(n+1)
=5-4(1/2)^n-(2n+1)(1/2)^n
=5-(1/2)^n (4+2n+1)
=5-(2n+5)*(1/2)^n
显然Sn

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