已知数列{an},a1=1,an>0且an+1^2=an^2/(4an^2+1),(n属于N+)

1个回答

  • 1.

    a(n+1)²=an²/(4an²+1)

    1/a(n+1)²=(4an²+1)/an²=1/an² +4

    1/a(n+1)²-1/an²=4,为定值

    1/a1²=1/1²=1,数列{1/an²}是以1为首项,4为公差的等差数列

    1/an²=1+4(n-1)=4n-3

    an>0

    an=1/√(4n-3)=√(4n-3)/(4n-3)

    数列{an}的通项公式为an=√(4n-3)/(4n-3)

    2.

    S(n+1)/an²=Sn/a(n+1)² +16n²-8n-3

    (4n-3)S(n+1)=[4(n+1)-3]Sn +(4n+1)(4n-3)

    等式两边同除以(4n+1)(4n-3)

    S(n+1)/[4(n+1)-3]=Sn/(4n-3) +1

    S(n+1)/[4(n+1)-3]-Sn/(4n-3)=1,为定值

    S1/(4×1-3)=b1/(4×1-3)=1/(4×1-3)=1

    数列{Sn/(4n-3)}是以1为首项,1为公差的等差数列

    Sn/(4n-3)=1+1×(n-1)=n

    Sn=n(4n-3)=4n²-3n

    n≥2时,

    bn=Sn-S(n-1)=4n²-3n-[4(n-1)²-3(n-1)]=8n-7

    n=1时,b1=8×1-7=1,同样满足通项公式

    数列{bn}的通项公式为bn=8n-7

    An=2×b1+2²×b2+2³×b3+...+2ⁿ×bn

    =2×(8×1-7)+2²×(8×2-7)+2³×(8×3-7)+...+2ⁿ×(8n-7)

    2An=2²×(8×1-7)+2³×(8×2-7)+...+2ⁿ×[8(n-1)-7]+2^(n+1)×(8n-7)

    An-2An=-An=2+8×2²+8×2³+...+8×2ⁿ -(8n-7)×2^(n+1)

    =8×(1+2+...+2ⁿ)-(8n-7)×2^(n+1) -22

    =8×1×[2^(n+1)-1]/(2-1) -(8n-7)×2^(n+1) -22

    =(15-8n)×2^(n+1) -30

    An=(8n-15)×2^(n+1) +30

    3.

    Tn=a1²+a2²+...+an²

    T(2n+1)-Tn=[a1²+a2²+...+a(2n+1)²]-(a1²+a2²+...+an²)

    =a(n+1)²+a(n+2)²+...+a(2n+1)²

    [T(2n+3)-T(n+1)]-[T(2n+1)-Tn]

    =[a(n+2)²+a(n+3)²+...+a(2n+1)²+a(2n+2)²+a(2n+3)²]-[a(n+1)²+a(n+2)²+...+a(2n+1)²]

    =a(2n+2)²+a(2n+3)²-a(n+1)²

    =1/[4(2n+2)-3] +1/[4(2n+3)-3] -1/[4(n+1)-3]

    =1/(8n+5) +1/(8n+9) -1/(4n+1)

    =1/(8n+5)+1/(8n+9)- 2/(8n+2)

    =[1/(8n+5)-1/(8n+2)]+[1/(8n+9)-1/(8n+2)]