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!

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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!
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!

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!
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

n是那个加法算式的项数,即有n个数相加。
第一问,找出一个值K,使得无论n取任何值,算式的和都不大于K。
第二问,n多大时,算式的和大于99%*K
解答
等比数列,公比是1/1.8
求和,得 108[1-(1/1.8)^n]/(1-1/1.8)
当n趋向无穷大,和为243
所以K值为243
第二问
243*99%=240.57...

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n是那个加法算式的项数,即有n个数相加。
第一问,找出一个值K,使得无论n取任何值,算式的和都不大于K。
第二问,n多大时,算式的和大于99%*K
解答
等比数列,公比是1/1.8
求和,得 108[1-(1/1.8)^n]/(1-1/1.8)
当n趋向无穷大,和为243
所以K值为243
第二问
243*99%=240.57

108[1-(1/1.8)^n]/(1-1/1.8)>240.57,n为正整数
求满足条件的n的最小值
得n=8

收起

就是求108+60+33 1/3+...等式的极限k
还有n为多少时,该等式=0.99*极限k
Q=1.8 Sn=a1[1-(1/Q)^n]/(1-1/Q)=108[1-(1/1.8)^n]/(1-1/1.8)=243=K
0.99K=108[1-(1/1.8)^n]/(1-1/1.8)
n=8