--- In

mathforfun@yahoogroups.com, "ramsey2879" <ramsey2879@...>

wrote:

>

> A.Let n go from 0 to 7

> B. Using the 3-bit binary values for 0 to 7, [000,001 ... 110,111]

> let (a,b,c)*n = (d,e,f) =(a-(4 and n)/4},b-(2 and n)/2,(c-(1 and

n))

>

> i.e.

> (2,3,4)*0 = (2,3,4)

> (2,3,4)*1 = (2,3,3)

> (2,3,4)*2 = (2,2,4)

> (2,3,4)*3 = (2,2,3)

> ...

> (2,3,4)*6 = (1,2,4)

> (2,3,4)*7 = (1,2,3)

>

> Let F(a,b,c) = T(a)*T(b)*T(c) where T(x) = x(x+1)/2 are the ordinary

> triangular numbers

>

> Prove that Sum (n = 0 to 7) F((a,b,c)*n) = (abc)^2

Yes, writing n as a three digit binary number n = b1b2b3:

(a,b,c)*n = (a-b1,b-b2,c-b3)

and the sum over all three digit binary numbers can be written as the

nested sum:

Sum (b1=0,1) Sum(b2=0,1) Sum(b3=0,1) F(a-b1,b-b2,c-b3)

=

Sum (b1=0,1) Sum(b2=0,1) Sum(b3=0,1) T(a-b1)T(b-b2)T(c-b3)

=

(T(a)+T(a-1))*(T(b)+T(b-1))*(T(c)+T(c-1))

but:

T(a)+T(a-1) = (a/2)(a+1 + a-1) = a^2 etc.. QED.