Any number that end in the digit x is raised to a power that ends in x+1.
The last digit of this number to the power is the same as that for x raised to the same power.
Each digit raised to successive powers has a cycle of length 1, 2 or 4. (For example the digit 3 has the cycle {3,9,7,1})
So it we look at the first 2000 terms of the sequence above, each last digit, whatever it may be will be included 2000, 1000, or 500 times. Therefore the last digit of
1
^{2} + 2
^{3} + 3
^{4} + 4
^{5} + .......+ 2000
^{2001} is zero.
So we really only need the last digits of the final 14 terms. By closer examination of the cycles I noted above the digits are of these final terms are 1, 8, 1, 4, 5, 6, 1, 8, 1, 0, 1, 2, 9, 4.
The final digit of this sum is
1.
Edited on July 14, 2014, 10:20 am

Posted by Jer
on 20140712 23:51:27 