已知数列{an}的前n项和为Sn

显示全部楼层 · 2008-8-25 22:06:52

已知数列{an}的前n项和为Sn,且对任意正整数n都有2Sn=(n+2)an-1 (1)求数列{an}的通项公式 (2)设Tn=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)…+1/(an*an+1),求Tn.

千问 · 2008-8-25 22:06:52

因为2Sn=(n+2)an-1n为任意正整数
2S(n-1)=(n+1)a(n-1)-1
2an=(n+2)an-(n+1)a(n-1)
nan-(n+1)a(n-1)=0
an/a(n-1)=(n+1)/n 2s1=3a1-1 s1=a1 a1=1 an/a1=an/a(n-1)*a(n-1)/a(n-2)....*a2/a1
=(n+1)/n*n/(n-1)......*3/2 an=(n+1)/2(2) Tn=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)…+1/(an*an+1)
=4/2*3+4/3*4+....4/(n+1)(n+2...

千问 · 2008-8-25 22:06:52

(1)因为s1=a1，所以2a1=3a1-1，得到a1=12Sn=(n+2)an-1 2Sn-1=(n+1)an-1-1 故2an=(n+2)an-(n+1)an-1 nan=(n+1)an-1 an=[(n+1)/n]an-1 =[(n+1)/n][n/(n-1)]an-2 =...= =[(n+1)/n][n/(n-1)]...

千问 · 2008-8-25 22:06:52

2a1=3a1-1 a1=1 2Sn=[(n+2)an]-1 2S（n-1）=[(n-1+2)a（n-1）]-1 2Sn-2S（n-1）=nan+2an-na（n-1）-a（n-1）=2an an/a（n-1）=（n+1）/n （a2/a1）×（a3/a2）×（a4/a3）×。。。。×[an/a（n-1）] =（3/2）×（4/3）×...