Show that .
We have three ways to prove this. The first proof is algebraic: we find an explicit expression in terms of n for both sides. So, we use Faulhaber’s formula. We recall
Thus, for the left-hand side, we have
Now, for the right-hand side, we have
and so the two are equal.
The second proof requires a little combinatorial thinking. We see from Faulhaber’s formula that the summand on the right is the kth triangular number:
So we can rewrite the right-hand sum as
Now, in the expansion on the right above, how many times does each number appear in the sum? We see that 1 occurs n times, once in all n inner sums; 2 occurs n-1 times, once in all n inner sums except the first; 3 occurs n-2 times, once in all n inner sums except the first and second; and so on. Thus, the value k, 1≤k≤n occurs n+1-k times in each sum, so combining all the k terms, we have the contribution of to the sum for each k, and so the total sum is , and we have proved our statement again.
The third proof comes from interpreting the sums geometrically. Consider a set of objects, such as spheres, packed into a tetrahedron with n per edge, like seen in the figure below for n=4:
If we slice this array parallel to a pair of opposing edges, the slices are rectangles, as done via the colors below
We see then that the rectangles go , and so the total number of objects is .
But we can also slice the tetrahedron parallel to a face, as done with the base in the figure below:
And we see we have triangles; the first n triangles, and so the number of objects is the sum of the first n triangular numbers:
and our third proof is complete.
Note that we have found that the nth tetrahedral number is .
Monday Math 138
Show that .