清空筛选条件
清空筛选条件

generate a summation series

5 次查看（过去 30 天）
Odien
Odien 2015-8-13
编辑： Walter Roberson 2021-6-12
3(2+1)+4(3+2+1)+5(4+3+2+1)+6(5+...1)+...+1000(999+...+1).
How to write a matlab code to calculate the following summation without using for loop?
i can see the trend for 3(2+1) if the n = 1 (n+2)[((n+2)-1)+((n+2)-2)].
Assume the summation only sum until 6th term.
can we use array to solve this?

请先登录，再回答此问题。

采纳的回答

Stephen23
Stephen23 2015-8-13
编辑：Stephen23 2015-8-13
>> X = 1+cumsum(2:999); % 1+[2,3+2,4+3+2,...,999+..+2]
>> Y = 3:1000; % [3,4,5,...,1000]
>> sum(X.*Y)
ans = 125083208248
Note that this uses element-wise multiplication.

更多回答（1 个）

Walter Roberson
Walter Roberson 2015-8-13
It does have a symbolic answer that you could find using nested symsum() if you have the symbolic toolbox.
But for a numeric answer: cumsum() and multiply by something gives a bunch of terms...

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by