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)!)

Replacing 2014 with smaller numbers led me to https://oeis.org/A001788
But this doesn't have much of a formula.

But I found one that I have not proven is correct (but it works as far as I've checked):
a(n)=n(n+1)2^(n-2)

a(2014)=2014*2015*2^2012
=4058210*2^2012

Since 2^2012 ends in 6, a(2014) ends in 60.

Posted by Jer on 2014-11-13 14:00:50