以知数列{an}的前n项和为Sn,且满足Sn=1/4(a+1)^2,an>0.1:
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2008-08-13
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S1=a1=1/4(a1+1)^2, 则4a1=(a1+1)^2, 推导得:(a1-1)^2=0, 则a1=1
S2=a1+a2=1/4(a2+1)^2, 则4+4a2=(a2+1)^2, 推导得:(a2-1)^2=4, 因a>0, 则a2=3
Sn-Sn-1=an=1/4(an+1)^2-1/4(`an-1`+1)^2,则4an=(an-`an-1`)(an+`an-1`+2), 推导得:(an+`an-1`)(an-`an-1`-2)=0, 因a>0,则an=`an-1`+2, 已知a1=1,则an=1+2(n-1)=2n-1
已知bn=20-an, 则bn=21-2n
若要求Sbn最大,则要求`Sn-1`<Sn<`Sn+1`, 则bn>0, `bn+1`<0
推导得:21-2n>0, 19-2n<0
则9.5<n<10.5, 且n为整数,那么n=10
S2=a1+a2=1/4(a2+1)^2, 则4+4a2=(a2+1)^2, 推导得:(a2-1)^2=4, 因a>0, 则a2=3
Sn-Sn-1=an=1/4(an+1)^2-1/4(`an-1`+1)^2,则4an=(an-`an-1`)(an+`an-1`+2), 推导得:(an+`an-1`)(an-`an-1`-2)=0, 因a>0,则an=`an-1`+2, 已知a1=1,则an=1+2(n-1)=2n-1
已知bn=20-an, 则bn=21-2n
若要求Sbn最大,则要求`Sn-1`<Sn<`Sn+1`, 则bn>0, `bn+1`<0
推导得:21-2n>0, 19-2n<0
则9.5<n<10.5, 且n为整数,那么n=10
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