已知a1=2,点(an,a(n+1))在函数f(x)=x的平方+2x的图像上,其中n=1,2,...

来源:百度知道 编辑:UC知道 时间:2024/09/20 07:11:33
(2)记bn=(1/an)+【1/a(n+2)】,求数列{bn}的前n项和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,...,
(2)记bn=(1/an)+[1/a(n+2)],求数列{bn}的前n项和Sn.

解:
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,...,
f(a1)=(a1)^2+2(a1)=f(2)=2^2+2*2=8=(a1)*(a1+2)=a2,
f(a2)=(a2)^2+2(a2)=f(8)=64+16=80=(a2)*(a2+2)=a3,
f(a3)=(a3)^2+2(a3)=f(80)=6400+160=6560=80*(80+2)==(a3)*(a3+2)=a4,
f(a4)=(a4)^2+2(a4)=f(6560)=6560*(6560+2)=(a4)*(a4+2)=a5,
......,
f(a(n-1))=(a(n-1))^2+2(a(n-1))=an,
f(an)=(an)^2+2(an)=a(n+1),
f(a(n+1))=(a(n+1))^2+2(a(n+1))=a(n+2).

(2)记bn=(1/an)+[1/a(n+2)],求数列{bn}的前n项和Sn.
b1=(1/a1)+[1/a3]=1/2+1/80=41/80,
b2=(1/a2)+[1/a4]=1/8+1/6560=83/6560,
b3=(1/a3)+[1/a5],
......,
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