一道英文数学题,看不懂啊,find the sum to n trems of the geometric series 108+_数学解答

一道英文数学题,看不懂啊,
find the sum to n trems of the geometric series 108+60+33 1/3+...if k is the least number which exceeds this sum for all values of n,find k.find also the least value of n for which the sum exceeds 99% of k.
thanks!

人气：296 ℃　时间：2020-02-03 14:43:40

解答

求等比数列108、60、33 1/3.前n项的和；如果k是大于这个数列中所有数的和的最小值,求k（即求数列和的上限）；寻找到一个n,使得数列中的前n项和,大于0.99*k.
等比数列为：108*（5/9）^(x-1).和是243-108*（5/9）^(n-1)*(5/4).
k为108*（1-5/9）=243.
即0.99*243