Exactly! Well done.Quote:
Originally posted by Nick
I noticed that derivative
of 1 + t + t^2 + t^3 + ... + t^n
is the 0 + 1 + 2t + 3t^2 +... + nt^(n-1)
I'm sure you could give a prove or solve it some how
If you sum the first series, using the standard geometric series result, you can then differentiate that result to get the sum of the second series, which is the same answer as you got.
Summing first series (using standard geometric series result):
1 + t + t^2 + ... + t^n = (1 - t^(n+1))/(1 - t)
d(1 + t + t^2 + ... + t^n)/dt = 1 + 2t + 3t^2 + ... + nt^(n - 1)
Which implies that:
1 + 2t + 3t^2 + ... + nt^(n - 1) = d((1 - t^(n+1))/(1 - t))/dt
= (1 - t^n + nt^(n + 1) - nt^n)/((1 - t)^2)
I think that's quite a nice little solution...