等差数列1.若两个等差数列{an
a1+a15 =a2+a14=。。。。=a7+a9 =2a8
==>S15 = 15a8
T15 =15b8
a8/b8=S15/T15 =(7*15+3)/(15+3) =108/18=6
2
(a1+2)/2 =根号(2S1) =根号(2a1)
==>(a1+2)^2 =8a1
an是正数组成的数列==>a1=2
(a2+2)/2 =根号(2S2)=根号[2(a2+ 2)]
==>(a2+2)^2 =8(a2+2)
==>(a2+2)(a2+2 -8)=0
an是正数组成的数列 ==>a2 =6
(a3+2)/2 =根号(2S3)=根号[2(a2+ 8)]
==>(a3+2)^2 =8...全部
a1+a15 =a2+a14=。。。。=a7+a9 =2a8
==>S15 = 15a8
T15 =15b8
a8/b8=S15/T15 =(7*15+3)/(15+3) =108/18=6
2
(a1+2)/2 =根号(2S1) =根号(2a1)
==>(a1+2)^2 =8a1
an是正数组成的数列==>a1=2
(a2+2)/2 =根号(2S2)=根号[2(a2+ 2)]
==>(a2+2)^2 =8(a2+2)
==>(a2+2)(a2+2 -8)=0
an是正数组成的数列 ==>a2 =6
(a3+2)/2 =根号(2S3)=根号[2(a2+ 8)]
==>(a3+2)^2 =8(a3+8)
an是正数组成的数列 ==>a3=10
2) 求数列{an}的通项公式
(an+2)/2 =根号(2Sn)
===>Sn =(an+2)^2/8 。
。。。(1)
S(n-1) =[a(n-1)+2)^2/8 。。。
(2)
(1)-(2)==>
an = {(an+2)^2 -[a(n-1)+2)^2]}/8
(an)^2 -[a(n-1)]^2 =4an+4a(n-1)
[an +a(n+1)][an -a(n+1)]=4[an +a(n+1)]
{an}是正数组成的数列
==>an +a(n+1)不等于0,两边除之
==> an -a(n+1) =4
{an}是公差为4的等差数列
a1 =2
==>通项公式an =2+(n-1)*4 =4n-2
。收起