More on Gaussian Sums

In my previous essay on Gaussian sums I stated that I was unable to find a way to infer the formula for the higher-order Gaussian sums directly. Thanks to a little serendipity I have changed that fact, though I must confess that I feel embarrassed that I was too blind originally to see the obvious solution staring me in the face.

We have defined the j-th order Gaussian
sum of N as

(Eq'n 1)

with the proviso that the first-order Gaussian
sum consists of the sum of all of the positive integers up to and including N.
With that information in mind we can work out S_{2}(N) by imagining that
we have set out markers representing the first-order Gaussian sums from S_{1}(1)
up to and including S_{1}(N), laying the markers out vertically with
each sum to the right of the previous sum.

Imagine having placed those markers on a
rectangular grid of squares. The array extends N squares along its base and S_{1}(N)
squares up its right side, thereby defining a rectangle containing NS_{1}(N)
squares. From the total number of squares in that rectangle we must subtract the
number of the unoccupied squares, so, beginning at the lower left corner and
taking horizontal lines of squares, we subtract 1x2, then 2x3, then 3x4, and so
on up to (N-1)xN squares. Thus we have

(Eq'n 2)

But (n+1)n = 2S_{1}(n), so we
have

(Eq'n 3)

We solve that readily for

(Eq'n 4)

Repeating that procedure with progressively higher order Gaussian sums enables us to apply the principle of induction to infer that, in general,

(Eq'n 5)

If we then begin with S_{0}(N) = N and
apply that equation repeatedly for j taking successive integer values, we find
that

(Eq'n 6)

That equation is identical to the equation that I induced to describe the elements of Pascal's triangle.

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