How about just writing down a few terms?
5^1 - 5^0
5^2 - 5^1
...
5^100 - 5^99
---------------
5^100 - 5^0
If you want to do it the hard way, you could note that
∑(a-b) = ∑a - ∑b
Evaluate each telescoping sum:
1) Sum from i = 1 to 100 of (5^i - 5^(i-1)).
I know that I could apply the (n(n+1))/2 summation rule to the 5^i part of the question and then plug in 100, but I am unsure what to do for the second half --> -5^(i-1)! Thank you for any help!
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