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

and

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 *k*th 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 *n*th tetrahedral number is .

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Tags: Faulhaber's Formula, Math, Monday Math

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