一道极限题··急求.在数列{an}中,a2=2,a6=10,且数列{√an-1}为等差数列(那1不是an的下标,但都在根号内.),bn=1/(an+2-an+1)(an+1-an) (这里1,2都是下标..),求当n趋向于无穷大时,b1+b2+b3+……+bn的极
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![一道极限题··急求.在数列{an}中,a2=2,a6=10,且数列{√an-1}为等差数列(那1不是an的下标,但都在根号内.),bn=1/(an+2-an+1)(an+1-an) (这里1,2都是下标..),求当n趋向于无穷大时,b1+b2+b3+……+bn的极](/uploads/image/z/14190723-27-3.jpg?t=%E4%B8%80%E9%81%93%E6%9E%81%E9%99%90%E9%A2%98%C2%B7%C2%B7%E6%80%A5%E6%B1%82.%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca2%3D2%2Ca6%3D10%2C%E4%B8%94%E6%95%B0%E5%88%97%7B%E2%88%9Aan-1%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%88%E9%82%A31%E4%B8%8D%E6%98%AFan%E7%9A%84%E4%B8%8B%E6%A0%87%2C%E4%BD%86%E9%83%BD%E5%9C%A8%E6%A0%B9%E5%8F%B7%E5%86%85.%EF%BC%89%2Cbn%3D1%2F%EF%BC%88an%2B2-an%2B1%EF%BC%89%EF%BC%88an%2B1-an%EF%BC%89+%28%E8%BF%99%E9%87%8C1%2C2%E9%83%BD%E6%98%AF%E4%B8%8B%E6%A0%87..%29%2C%E6%B1%82%E5%BD%93n%E8%B6%8B%E5%90%91%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%E6%97%B6%2Cb1%2Bb2%2Bb3%2B%E2%80%A6%E2%80%A6%2Bbn%E7%9A%84%E6%9E%81)
一道极限题··急求.在数列{an}中,a2=2,a6=10,且数列{√an-1}为等差数列(那1不是an的下标,但都在根号内.),bn=1/(an+2-an+1)(an+1-an) (这里1,2都是下标..),求当n趋向于无穷大时,b1+b2+b3+……+bn的极
一道极限题··急求.
在数列{an}中,a2=2,a6=10,且数列{√an-1}为等差数列(那1不是an的下标,但都在根号内.),bn=1/(an+2-an+1)(an+1-an) (这里1,2都是下标..),求当n趋向于无穷大时,b1+b2+b3+……+bn的极限
蛋疼·。答案是2/3···哪个能证明8/3是对的·答案是错的·?或者给个确定的理由·。
一道极限题··急求.在数列{an}中,a2=2,a6=10,且数列{√an-1}为等差数列(那1不是an的下标,但都在根号内.),bn=1/(an+2-an+1)(an+1-an) (这里1,2都是下标..),求当n趋向于无穷大时,b1+b2+b3+……+bn的极
{√an-1}为等差数列
{√a2-1}=1
{√a6-1}=3
因此每项之间差为(3-1)/4 =0.5
(√an-1) = (√a2-1) + 0.5 * (n-2) =0.5n
an = (0.5n)^2+1
a(n+2)-a(n+1) = (0.5n+1)^2-(0.5n+0.5)^2 = 0.5(n+1.5)
a(n+1)-a(n) = (0.5n+0.5)^2 -(0.5n)^2 = 0.5(n+0.5)
bn = 1/((n+3/2)(n+1/2)/4) = 16/[(2n+3)(2n+1)]
= 8[1/(2n+1) -1/(2n+3)]
Sn=b1+b2+b3+...+bn = 8[1/3-1/5+1/5-1/7+.+1/(2n+1) -1/(2n+3)]
=8[1/3-1/(2n+3)]
极限是8/3
bn=1/(an+2-an+1)(an+1-an) 是 bn=1/ [(an+2-an+1)(an+1-an)] 吗?
√(a6-1)=3 √(a2-1)=1
a'n=√(an-1)是等差数列,a'2=1,a'6=3
a'2=a'1+d=1
a'6=a'1+5d=3
a1'=1/2 d=1/2 √(a1-1)=1/2 a1=5/4
an'=(1/2)+(n-1)/2=n/2
a'n-1=(n-1)/2
an-an-1=[√(an-1)]^2-[...
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√(a6-1)=3 √(a2-1)=1
a'n=√(an-1)是等差数列,a'2=1,a'6=3
a'2=a'1+d=1
a'6=a'1+5d=3
a1'=1/2 d=1/2 √(a1-1)=1/2 a1=5/4
an'=(1/2)+(n-1)/2=n/2
a'n-1=(n-1)/2
an-an-1=[√(an-1)]^2-[√(an-1 -1)]^2=n^2/4-(n-1)^2/4=(2n-1)/4
Sn-a1-Sn-1=[(3/4)+(2n-1)/4](n-1)/2 an=5/4+(n-1)^2/4
an+2-an+1=(n+1)^2/4-n^2/4=(2n+1)/4
an+1-an=n^2/4-(n-1)^2/4=(2n-1)/4
bn=16/[(2n-1)(2n+1)]=8*[1/(2n-1)-1/(2n+1)]
b1=8*(1-1/3)
b1+b2+..+bn=8*[1-1/(2n+1)]
lim(n→+∞)b1+b2+b3+..+bn=8
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