[(2³-1﹚﹙3³﹣1﹚.﹙100³-1﹚]÷[﹙2³+1﹚﹙3³+1﹚……﹙100³+1﹚]的值最接近多少
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 21:58:38
![[(2³-1﹚﹙3³﹣1﹚.﹙100³-1﹚]÷[﹙2³+1﹚﹙3³+1﹚……﹙100³+1﹚]的值最接近多少](/uploads/image/z/10780040-56-0.jpg?t=%5B%EF%BC%882%26%23179%3B%EF%BC%8D1%EF%B9%9A%EF%B9%993%26%23179%3B%EF%B9%A31%EF%B9%9A.%EF%B9%99100%26%23179%3B%EF%BC%8D1%EF%B9%9A%5D%C3%B7%5B%EF%B9%992%26%23179%3B%2B1%EF%B9%9A%EF%B9%993%26%23179%3B%EF%BC%8B1%EF%B9%9A%E2%80%A6%E2%80%A6%EF%B9%99100%26%23179%3B%EF%BC%8B1%EF%B9%9A%5D%E7%9A%84%E5%80%BC%E6%9C%80%E6%8E%A5%E8%BF%91%E5%A4%9A%E5%B0%91)
[(2³-1﹚﹙3³﹣1﹚.﹙100³-1﹚]÷[﹙2³+1﹚﹙3³+1﹚……﹙100³+1﹚]的值最接近多少
[(2³-1﹚﹙3³﹣1﹚.﹙100³-1﹚]÷[﹙2³+1﹚﹙3³+1﹚……﹙100³+1﹚]的值最接近多少
[(2³-1﹚﹙3³﹣1﹚.﹙100³-1﹚]÷[﹙2³+1﹚﹙3³+1﹚……﹙100³+1﹚]的值最接近多少
100很大,你可以设通项为(n^3-1)/(n^3+1),原式为n从2到100通项的乘积
∏(n^3-1)/(n^3+1)
=∏(n-1)[n(n+1)+1]/{(n+1)[(n-1)n+1]}
=∏[(n-1)/(n+1)]*∏{[n(n+1)+1]/[(n-1)n+1]}
={2/[n(n+1)]}{[n(n+1)+1]/(1*2+1)}
=(2/3){[n(n+1)+1]/[n(n+1)]}
[n(n+1)+1]/[n(n+1)]当n很大时,趋于1
∴(2/3){[n(n+1)+1]/[n(n+1)]}趋于2/3
最接近的话 ,应该是1