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)]
