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20

∑ k

k=10

Normally, I would think to use the theorem for the sum of the first n integers:

n

∑ k = n(n+1)/2

k=1

I don't know how to do this, however, since the lower limit is k=10, not k=1.

My professor wrote this note on the board for the problem. It gives the answer, but I still don't understand it. If you do, could you explain? Thanks!

20....20....9

∑ k = ∑ - ∑ k

k=10 k=1 k=1

20(21)/2 - 9(10)/2 = 165 <-- (The answer)