Problem: There are $$$n$$$ sticks of length $$$1,2,3,...,n$$$. How many incongruent triangles can be formed by using three of the given sticks?
My solution: After solving the problem, I found the number of triangles, $$$x$$$, to be:
However, I am not satisfied with this expression. Can it be simplified? (Can we express this equation without using the summation sign?)
Notice that there are 2 cases for $$$\lceil\frac{i+1}2\rceil$$$. It is either $$$\frac{i+1}2$$$, either $$$\frac{i+2}2$$$, depending on $$$i$$$'s parity. You should try to break down the sums into odd sums and even sums and get an answer that way.
Let $$$n = 2k+1$$$ for some integer $$$k$$$
For odd $$$i\ (i = 2m-1)$$$
For even $$$i\ (i = 2m)$$$
Therefore, if $$$n$$$ is odd
And if n is even
Notice that:
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Finally,
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Similarly for
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Let
Then
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Final one:
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Let
Then
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Finally, if $$$n$$$ is odd:
if $$$n$$$ is even: