# Riddles/Math/Logic

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• 09-03-2002
rahaydenuk
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
using this?

Exactly! Well done.

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.

Like this:

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...
• 09-03-2002
Nick
If you havn't got it already check out
concrete mathematic --- A foundation for computer science.
Lots of neat problems, most of them are solved in the back
of book.

I'm trying to read this book in my spare time, but I have
lots of course work so I havn't gotten too far. It's tough