Difference between revisions of "Talk:Algo-analysis-TADM2E"
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(Created page with "Hello, I have a solution to the task 2-34 (TADM2E 2.34) but I have no permissions to created/edit anything. Here is the solution: This task refers to this song The_Twelve_Da...") |
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Here is the solution: | Here is the solution: | ||
| − | This task refers to this song The_Twelve_Days_of_Christmas_(song) | + | This task refers to this song The_Twelve_Days_of_Christmas_(song). |
| + | |||
So we have to calculate the sum like this | So we have to calculate the sum like this | ||
1 day - 1 gift | 1 day - 1 gift | ||
| + | |||
2 day - 1 gift + 2 gifts | 2 day - 1 gift + 2 gifts | ||
| + | |||
3 day - 1 + 2 + 3 | 3 day - 1 + 2 + 3 | ||
| + | |||
n day - 1 + 2 + 3 + ... + n | n day - 1 + 2 + 3 + ... + n | ||
Revision as of 09:31, 26 May 2020
Hello, I have a solution to the task 2-34 (TADM2E 2.34) but I have no permissions to created/edit anything.
Here is the solution:
This task refers to this song The_Twelve_Days_of_Christmas_(song).
So we have to calculate the sum like this
1 day - 1 gift
2 day - 1 gift + 2 gifts
3 day - 1 + 2 + 3
n day - 1 + 2 + 3 + ... + n
Formula is
$ \begin{align} &\sum_{i=1}^n \sum_{j=1}^n j\ = \sum_{i=1}^n \frac{n(n+1)}{2} =\\ &= \frac{1}{2}\sum_{i=1}^n n^2 + \frac{1}{2}\sum_{i=1}^n n=\\ &= \frac{n(n+1)(2n+1) + 3n(n+1)}{12} = \frac{(n+1)(n^2+2n)}{6}\ \end{align} $
