急急急已知函数f(x)=1/x,数列an的前n项和为sn,点Pn(an^2,1/(an+1)^2-4)都在函数f(x)的图像上且a1=1,an>0(1)求an通项公式(2)若数列bn的前n 项和为Tn且满足Tn+1/an的平方=Tn/an+1的平方+(4n-3)(4n+1)试
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![急急急已知函数f(x)=1/x,数列an的前n项和为sn,点Pn(an^2,1/(an+1)^2-4)都在函数f(x)的图像上且a1=1,an>0(1)求an通项公式(2)若数列bn的前n 项和为Tn且满足Tn+1/an的平方=Tn/an+1的平方+(4n-3)(4n+1)试](/uploads/image/z/2681991-63-1.jpg?t=%E6%80%A5%E6%80%A5%E6%80%A5%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2Fx%2C%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E7%82%B9Pn%EF%BC%88an%5E2%2C1%2F%28an%2B1%29%5E2-4%EF%BC%89%E9%83%BD%E5%9C%A8%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8A%E4%B8%94a1%3D1%2Can%3E0%EF%BC%881%EF%BC%89%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn+%E9%A1%B9%E5%92%8C%E4%B8%BATn%E4%B8%94%E6%BB%A1%E8%B6%B3Tn%2B1%2Fan%E7%9A%84%E5%B9%B3%E6%96%B9%3DTn%2Fan%2B1%E7%9A%84%E5%B9%B3%E6%96%B9%2B%EF%BC%884n-3%EF%BC%89%EF%BC%884n%2B1%29%E8%AF%95)
急急急已知函数f(x)=1/x,数列an的前n项和为sn,点Pn(an^2,1/(an+1)^2-4)都在函数f(x)的图像上且a1=1,an>0(1)求an通项公式(2)若数列bn的前n 项和为Tn且满足Tn+1/an的平方=Tn/an+1的平方+(4n-3)(4n+1)试
急急急已知函数f(x)=1/x,数列an的前n项和为sn,点Pn(an^2,1/(an+1)^2-4)都在函数f(x)的图像上且a1=1,
an>0(1)求an通项公式(2)若数列bn的前n 项和为Tn且满足Tn+1/an的平方=Tn/an+1的平方+(4n-3)(4n+1)试确定b1的值,使得bn 是等差数列
那个 的平方 没有包括Tn
急急急已知函数f(x)=1/x,数列an的前n项和为sn,点Pn(an^2,1/(an+1)^2-4)都在函数f(x)的图像上且a1=1,an>0(1)求an通项公式(2)若数列bn的前n 项和为Tn且满足Tn+1/an的平方=Tn/an+1的平方+(4n-3)(4n+1)试
(1)将Pn代入f(x)得到
1/(an+1)^2-4=1/an^2
1/(an+1)^2-1/an^2=4
所以1/an^2是等差数列
1/an^2=1/a1^2+4*(n-1)=4*n-3
an>0,所以an=1/根号(4*n-3)