More on Gaussian Sums

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    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 S2(N) by imagining that we have set out markers representing the first-order Gaussian sums from S1(1) up to and including S1(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 S1(N) squares up its right side, thereby defining a rectangle containing NS1(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 = 2S1(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 S0(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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