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Deux Digit Determination (
Posted on 20141113
)
Find the last two digits in the base ten expansion of :
C(2014, 1) + 2
^{2}
*C(2014, 2) + 3
^{2}
*C(2014, 3) + ......+ 2014
^{2}
*C(2014, 2014)
**** C is the choose function, that is:
C(m,n) = m!/(n!*(mn)!)
See The Solution
Submitted by
K Sengupta
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Full expansion, but unproven result
 Comment 1 of 4
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^(n2)
a(2014)=2014*2015*2^2012
=4058210*2^2012
Since 2^2012 ends in 6, a(2014) ends in 60.
Posted by
Jer
on 20141113 14:00:50
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