Find the last two digits in the base ten expansion of :

C(2014, 1) + 2²*C(2014, 2) + 3²*C(2014, 3) + ......+ 2014²*C(2014, 2014)

**** C is the choose function, that is:
C(m,n) = m!/(n!*(m-n)!)

(In reply to Full expansion, but unproven result by Jer)

Calculating 4058210*2^2012 in UBASIC (under DOSBox) does come out equal to the number calculated directly in my computer solution.

Posted by Charlie on 2014-11-13 14:29:50