"Counting in their heads" - 1895 oil painting

[Albert Chan]

(08-08-2020 07:42 PM)Albert Chan Wrote:  \(F(x) = \sum_{t=0}^{x-1} f(t) = 100\binom{x}{1} + 21\binom{x}{2} + 2\binom{x}{3}\)

F(5)/365 = (100*5 + 21*10 + 2*10) / 365 = 730/365 = 2

Another way to get F(x), via Triangular numbers

\(\sum_{x=1}^n \binom{x+0}{1} = \binom{n+1}{2} \)           // triangular numbers
\(\sum_{x=1}^n \binom{x+1}{2} = \binom{n+2}{3} \)           // sum of triangular numbers

10² + 11² + 12² + 13² + 14²
= 10² + (10²+21) + (10²+21+23) + (10²+21+23+25) + (10²+21+23+25+27)
= 100*5 + 21*4 + 23*3 + 25*2 + 27*1
= 100*5 + 21*(4+3+2+1) + 2*(1+1+1+2+2+3)
= 100*5 + 21*(1+2+3+4) + 2*(1+(1+2)+(1+2+3))

= \(100\binom{5}{1} + 21\binom{5}{2} + 2\binom{5}{3}\)

This matched the quoted F(x) equation, thus gives the same answer