from a geometric progression is given t3(角标3)=162 and t6=6,calculate the sumlimit S
问题描述：
from a geometric progression is given t3(角标3)=162 and t6=6,calculate the sumlimit S
(大体就是告诉等比数列的两项,可是那个sumlimit S是什么玩意啊?我百度了,可是查不到...)
答
【题目】
等比数列中t3=162,t6=6,求数列和的极限S.
设公比为q,则q³=t6/t3=1/27
==> q=1/3
==> 首项 t1=t3/q²=1458
==> 前n项和 Sn=t1•(1-q^n)/(1-q)
当n→∞时,Sn→S=t1/(1-q)=2187
sum：和
limit：极限
sumlimit就是数列前n项和的极限.