Summation on HP 42S

[Albert Chan]

(09-24-2018 12:56 AM)Thomas Klemm Wrote:  The summation leads to:

\(\sum_{x=1}^{n} 2\binom{x}{2} - 2\binom{x}{1} = 2\binom{n+1}{3} - 2\binom{n+1}{2} = 2\left ( \binom{n+1}{3}-\binom{n+1}{2} \right )\)

This almost look like integration !
Any reference to show it is always true ?

I normally just fit the data to search for the formula (assume I don't cheat by googling)
For sum(x^2 - 3*x), formula should be a cubic, with no constant term (sum=0 when n=0)

Try:
n = 1, sum = (1^2 - 3) = -2
n = 2, sum = -2 + (2^2 - 3*2) = -4
n = 3, sum = -4 + (3^2 - 3*3) = -4

3 equations, 3 unknowns (cubic coefficients), we get sum = n^3/3 - n^2 - 4/3*n

So, for n = 100, sum = n/3 * (n^2 - 3*n - 4) = 100/3 * 9696 = 323200