数列求和问题

显示全部楼层 · 2008-6-6 14:07:02

通项公式=3/(6n-5)(6n+1)
然后求和=3/1x7+3/7x13+3/13x19+.....+3/(6n-5)(6n+1)
我不懂怎么化简上面那个长长的式子
答案= 1/2[1-1/(6n+1)]

千问 · 2008-6-6 14:07:02

这类题目都是先研究通项工式an=3/(6n-5)(6n+1)=1/2[1/(6n-5)-1/(6n+1)]2an=1/(6n-5)-1/(6n+1)2a1=1-1/72a2=1/7-1/132a3==1/13-1/19...2an=1/(6n-5)-1/(6n+1)两边相加2sn=1-1/(6n+1)sn=1/2[1-1/(6n+1)]你的解法和我和解法是一样的,只不过把系数放的地方不一样而已.

千问 · 2008-6-6 14:07:02

通项公式=3/(6n-5)(6n+1)=1/2[1/(6n-5)-1/(6n+1)] 求和=3/1x7+3/7x13+3/13x19+.....+3/(6n-5)(6n+1) =1/2[1/1-1/7]+1/2[1/7-1/13]+...+1/2[1/(6n-5)-1/(6n+1)]=1/2[1-1/7+1/7-1/13+....+1

千问 · 2008-6-6 14:07:02

3/1x7+3/7x13+3/13x19+.....+3/(6n-5)(6n+1) =0.5*(6/1x7+6/7x13+6/13x19+.....+6/(6n-5)(6n+1))=0.5(1-1/7+1/7-1/13+1/13-1/19+...+1/(6n-5)-1/(6n+1))=0.5(1-1/(6n+1))=3n/(6n+1)

千问 · 2008-6-6 14:07:02

你看啊3/(6n-5)(6n+1) ＝0.5*[1/(6n-5)-1/(6n+1)]然后带入中间的全消了只有首项减尾项

千问 · 2008-6-6 14:07:02

求和=1/2[1-1/7+1/7-1/13......1/(6n-5)-1/(6n+1)]
=1/2[1-1/(6n+1)]