问一道关于数列的题设正项数列{an}的前n项和为Sn,并且对于任意n∈N*,an与1的等差中项等于√Sn,求数列{an}的通项公式.急用,
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问一道关于数列的题设正项数列{an}的前n项和为Sn,并且对于任意n∈N*,an与1的等差中项等于√Sn,求数列{an}的通项公式.急用,
问一道关于数列的题
设正项数列{an}的前n项和为Sn,并且对于任意n∈N*,an与1的等差中项等于√Sn,求数列{an}的通项公式.
急用,
问一道关于数列的题设正项数列{an}的前n项和为Sn,并且对于任意n∈N*,an与1的等差中项等于√Sn,求数列{an}的通项公式.急用,
an+1=2√Sn
令n=1,解得a1=1
平方得,an²+2an+1=4Sn
当n≥2时,a(n-1)²+2a(n-1)+1=4Sn-1
两式相减得,an²-a(n-1)²+2[an-a(n-1)]=4an
an²-a(n-1)²=2[an+a(n-1)]
整理得:an-a(n-1)=2
所以{an}是等差数列,首项为1,公差为2
所以通项为.
an+1=2√Sn
(an+1)^2=4Sn
[a(n+1)+1]^2=4S(n+1)
[a(n+1)+1]^2-(an+1)^2=4S(n+1)-4Sn
[a(n+1)^2+2a(n+1)+1]-(an+1)^2=4a(n+1)
[a(n+1)^2-2a(n+1)+1]-(an+1)^2=0
[a(n+1)-1]^2-(an+1)^2=0
...
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an+1=2√Sn
(an+1)^2=4Sn
[a(n+1)+1]^2=4S(n+1)
[a(n+1)+1]^2-(an+1)^2=4S(n+1)-4Sn
[a(n+1)^2+2a(n+1)+1]-(an+1)^2=4a(n+1)
[a(n+1)^2-2a(n+1)+1]-(an+1)^2=0
[a(n+1)-1]^2-(an+1)^2=0
{[a(n+1)-1]+(an+1)}*{[a(n+1)-1]-(an+1)}=0
[a(n+1)+an]*[a(n+1)-an-2]=0
a(n+1)=－an 或者 a(n+1)=an+2
(a1+1)^2=4S1=4a1
a1=1
因为数列是正数列，所以a(n+1)=－an应该舍去
an＝1＋（n－1）2＝2n－1
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