An=根号Sn+根号(Sn+1) 证明根号Sn是等差数列,求An通项公式
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An=根号Sn+根号(Sn+1) 证明根号Sn是等差数列,求An通项公式
应该是
a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2)...
s(n+1)-s(n)=a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2),
{[s(n+1)]^(1/2)-[s(n)]^(1/2)}{[s(n+1)]^(1/2)+[s(n)]^(1/2)}=[s(n)]^(1/2)+[s(n+1)]^(1/2),
[s(n+1)]^(1/2)-[s(n)]^(1/2)=1,
{[s(n)]^(1/2)}是首项为[s(1)]^(1/2)=[a(1)]^(1/2),公差为1的等差数列.
[s(n)]^(1/2)=[a(1)]^(1/2) +(n-1),
a(n+1)=[s(n+1)]^(1/2) + [s(n)]^(1/2) = [a(1)]^(1/2) + n + [a(1)]^(1/2) +n-1
=2[a(1)]^(1/2) + 2(n+1) -3,
a(n)=2[a(1)]^(1/2)+2n-3,
a(1)=a^2,
a^2=a(1)=2a-1,
0=a^2-2a+1=(a-1)^2,
a=1.
a(n)=2+2n-3=2n-1
a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2)...
s(n+1)-s(n)=a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2),
{[s(n+1)]^(1/2)-[s(n)]^(1/2)}{[s(n+1)]^(1/2)+[s(n)]^(1/2)}=[s(n)]^(1/2)+[s(n+1)]^(1/2),
[s(n+1)]^(1/2)-[s(n)]^(1/2)=1,
{[s(n)]^(1/2)}是首项为[s(1)]^(1/2)=[a(1)]^(1/2),公差为1的等差数列.
[s(n)]^(1/2)=[a(1)]^(1/2) +(n-1),
a(n+1)=[s(n+1)]^(1/2) + [s(n)]^(1/2) = [a(1)]^(1/2) + n + [a(1)]^(1/2) +n-1
=2[a(1)]^(1/2) + 2(n+1) -3,
a(n)=2[a(1)]^(1/2)+2n-3,
a(1)=a^2,
a^2=a(1)=2a-1,
0=a^2-2a+1=(a-1)^2,
a=1.
a(n)=2+2n-3=2n-1
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