Figure out the pattern - part II

Employing Fubini Theorem, the nth term is defined by the following: $\displaystyle\sum_{k=1}^{n} \;\dfrac{k(k+1)}{2}=\displaystyle\sum_{k=1}^{n} \displaystyle\sum_{j=1}^{k} \;j=\displaystyle\sum_{j=1}^{n} \displaystyle\sum_{k=j}^{n} \;j=\displaystyle\sum_{j=1}^{n} \;j(n+1-j)=\displaystyle\sum_{k=1}^{n} \; k(n+1-k)=\displaystyle\sum_{k=1}^{n} \; k((n+2)-(k+1))$

$\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=(n+1)\displaystyle\sum_{k=1}^{n} \; k-2\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=(n+2)\dfrac{n(n+1)}{2}-2\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=\dfrac{n(n+1)(n+2)}{2}-2\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}$

$(2+1) \displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=\dfrac{n(n+1)(n+2)}{2}$

$3\;\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=\dfrac{n(n+1)(n+2)}{2}$

$\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=\dfrac{n(n+1)(n+2)}{3\times2}$

$\displaystyle\sum_{k=1}^{n} \; \dfrac{k(k+1)}{2}=\dfrac{n(n+1)(n+2)}{6}$

Thus we are seeking the 8th term which is $\dfrac{8(9)(10)}{6}=\boxed{120}$ . In fact, we can generalise the nth term to be $\dbinom{n}{3}$ .

You know triangular numbers. A good way of packing circles of uniform size is to build triangles. An efficient way of stacking spheres of uniform size or cannonballs is to build regular tetrahedrons. That's how we get the three dimensional triangular numbers. The n-th three dimensional triangular number equals the sum of the first n two dimensional triangular numbers. We can use this fact to derive the formula for the n-th three dimensional triangular number and the result is $\frac{n(n+1)(n+2)}{6}$ . In the given sequence those are the first several three dimensional triangular numbers. Therefore the next one should be 120. By the way triangular numbers of all dimensions lie on the diagonals in Pascal's triangle.