请将这个数列变成一个固定公式吧 6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]+6000x[1+(0.002x4天)]+6000x[1+(0.002x5天)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 21:03:51
![请将这个数列变成一个固定公式吧 6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]+6000x[1+(0.002x4天)]+6000x[1+(0.002x5天)]](/uploads/image/z/12507251-59-1.jpg?t=%E8%AF%B7%E5%B0%86%E8%BF%99%E4%B8%AA%E6%95%B0%E5%88%97%E5%8F%98%E6%88%90%E4%B8%80%E4%B8%AA%E5%9B%BA%E5%AE%9A%E5%85%AC%E5%BC%8F%E5%90%A7+6000x%5B1%2B%280.002x1%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x2%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x3%E5%A4%A9%EF%BC%89%5D6000x%5B1%2B%EF%BC%880.002x1%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x2%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x3%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x4%E5%A4%A9%EF%BC%89%5D%2B6000x%5B1%2B%EF%BC%880.002x5%E5%A4%A9%EF%BC%89%5D)
请将这个数列变成一个固定公式吧 6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]+6000x[1+(0.002x4天)]+6000x[1+(0.002x5天)]
请将这个数列变成一个固定公式吧 6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]
6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]+6000x[1+(0.002x4天)]+6000x[1+(0.002x5天)]……
6000是钱,也可用代替,天数是N,我想得到一个公式可以算出任意累计N天的和,谁可以帮帮我?
请将这个数列变成一个固定公式吧 6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]6000x[1+(0.002x1天)]+6000x[1+(0.002x2天)]+6000x[1+(0.002x3天)]+6000x[1+(0.002x4天)]+6000x[1+(0.002x5天)]
考察一般的加法因子第k个加法因子:
6000×(1+0.002×k)=6000+6000×0.002k=6000+12k
n天的和:
Sn=6000n+12(1+2+...+n)=6000n+12n(n+1)/2=6n²+6006n
得到公式:Sn=6n²+6006n,可以累计任意n天的和.