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...)
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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...)
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...)
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...)
【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.
sum limit就是这个geometric progression的极限,因为是递减所以有一个极限
ax^2=162
ax^6=6
x=1/3
a=1458
ax^n极限是1458/(3^n),应该是0吧
at a pro-rated
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...)
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