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)!)

Using the binomial expansion:

            Sum from r=0 to n of [ C(n,r)x^r ] = (1 + x)ⁿ

Differentiate wrt x:  Sum[ r*C(n,r)x^r-1 ] = n(1 + x)^n-1

Multiply by x:          Sum[ r*C(n,r)x^r ] =  nx(1 + x)^n-1

Differentiate wrt x:        

  Sum[ r²*C(n,r)x^r-1] = nx(n – 1)(1 + x)^n-2 + n(1 + x)^n-1

Substitute x = 1 and simplify:

             Sum[ r²*C(n,r) ] = n(n + 1)*2^n-2

Now, using n = 2014 and noticing that the first term in
the sum is zero (since r=0), we get

  Sum, r=1 to 2014, [ r²*C(2014,r) ] = 2014*2015*2²⁰¹²

with final digits 60, as Jer suggested.

Posted by Harry on 2014-11-16 17:22:22