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I am looking for the sum of consecutive odd triangular numbers. I am trying to relate the number of $k \times k$ rhombi in an $n \times n \times n$ equilateral triangle. While I have figured out an answer to the problem as it relates to sums of triangular numbers, in particular the sum of odd triangular numbers when n is even and the sum of even triangular numbers when $n$ is odd, I am having trouble finding a closed form for the expression. For example, in a $6 \times 6 \times 6$ equilateral triangle composed of unit triangles there are $66$ total rhombi; 1 $3 \times 3$, 6 $2 \times 2$, and 15 $1 \times 1$, and we have that $T_1=1$, $T_3=6$, and $T_5=15$. $1+6+15 = 22$. And there are three types of rhombi; left, right, and vertical, so $22 \times 3$ is $66$. The same works for $n=7$ but it is the sum of $T_2 + T_4 + T_6$.

So i know how to answer the question of total rhombi, but I can not get a closed form of the summation of odd or even triangular numbers.

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    $\begingroup$ By odd triangular numbers do you mean $1+2+\cdots+n$ for $n$ odd or do you mean triangular numbers which are odd? $\endgroup$ – anon Mar 25 '13 at 14:32
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$T_n=\dfrac{n(n+1)}{2}$

$\sum_{k=0}^n T_{2k+1}=\sum_{k=0}^n(2k+1)(k+1)$

Expand, and use the fact that $\sum_{k=0}^nk^2=\dfrac{k(k+1)(2k+1)}{6}$ and $\sum_{k=0}^nk=\dfrac{k(k+1)}{2}$

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  • $\begingroup$ wow....that was way easier that I thought....slightly embarassed... $\endgroup$ – Eleven-Eleven Mar 25 '13 at 14:46
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An automatic solution with GP:

T(n)=n*(n+1)/2
sumformal(T(2*n+1))

Alternately: odd triangular numbers are represented by a quadratic polynomial, and so their sum is a cubic polynomial. Work out the first four sums then interpolate.

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