已知a1=2,点(an,an+1)在函数f(x)=x^2+2x的图象上,其中n=1,2,3…记bn=1/an+1/(an+2),求数列{bn}的前n项和为Sn并证明Sn
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![已知a1=2,点(an,an+1)在函数f(x)=x^2+2x的图象上,其中n=1,2,3…记bn=1/an+1/(an+2),求数列{bn}的前n项和为Sn并证明Sn](/uploads/image/z/6913079-71-9.jpg?t=%E5%B7%B2%E7%9F%A5a1%3D2%2C%E7%82%B9%28an%2Can%2B1%29%E5%9C%A8%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2%2B2x%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A%2C%E5%85%B6%E4%B8%ADn%3D1%2C2%2C3%E2%80%A6%E8%AE%B0bn%3D1%2Fan%2B1%2F%28an%2B2%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E5%B9%B6%E8%AF%81%E6%98%8ESn)
已知a1=2,点(an,an+1)在函数f(x)=x^2+2x的图象上,其中n=1,2,3…记bn=1/an+1/(an+2),求数列{bn}的前n项和为Sn并证明Sn
已知a1=2,点(an,an+1)在函数f(x)=x^2+2x的图象上,其中n=1,2,3…
记bn=1/an+1/(an+2),求数列{bn}的前n项和为Sn并证明Sn<1
已知a1=2,点(an,an+1)在函数f(x)=x^2+2x的图象上,其中n=1,2,3…记bn=1/an+1/(an+2),求数列{bn}的前n项和为Sn并证明Sn
已知a1=2,点(an,a(n+1))在函数f(x)=x^2+2x的图像上,其中n=1,2,...,
则f(an)=(an)^2+2(an)=a(n+1),
显然an>0,a(n+1)=an(an+2).
bn=(1/an)+[1/a(n+2)]
数列{bn}的前n项和Sn
Sn=b1+b2+b3+.+bn=
=(1/a1+1/a3)+(1/a2+1/a4)+(1/a3+1/a5)+...+[1/an+1/a(n+2)]=
={1/a1+1/[(a2)*(a2+2)]}+{1/[(a1)*(a1+2)]+1/[(a3)*(a3+2)]}+.+{1/[(a(n-1))^2+2(a(n-1))]+1/[(a(n+1))^2+2(a(n+1))]},
因为 1/[(a1)*(a1+2)]=(1/2)[1/(a1)-1/(a3)],
1/[(a2)*(a2+2)]=(1/2)[1/(a2)-1/(a4)],
1/[(a3)*(a3+2)]=(1/2)[1/(a3)-1/(a5)],
.,
1/[(a(n-1))*(a(n-1)+2)]=(1/2)[1/(a(n-1)-1/(a(n+1))],
1/[(an))*(an+2)]=(1/2)[1/(an)-1/(a(n+2))],
1/[(a(n+1))*(a(n+1)+2)]=(1/2)[1/(a(n+1)-1/(a(n+3))],
所以,
Sn=1/a1+(1/2)[1/(a2)-1/(a4)]+(1/2)[1/(a1)-1/(a3)]+
+(1/2)[1/(a3)-1/(a5)]+.+(1/2)[1/(a(n-1)-1/(a(n+1))]+
+(1/2)[1/(a(n+1)-1/(a(n+1))]
=(3/2)(1/a1)+(1/2)(1/a2)-(1/2)[1/(a(n+3))]
=(3/2)(1/2)+(1/2)(1/8)-(1/2)[1/(a(n+3))]
=13/16-(1/2)[1/(a(n+3))]