Re: Number with a_n..


"Timothy Murphy" <gayleard@xxxxxxxxxx> wrote in message
news:mob2k.25727$j7.470124@xxxxxxxxxxxxxxxxx
mina_world wrote:

My question is...
1.2 + 2. 2^2 + 3.2^3 + ... + (n-2).2^(n-2) = ?
Namely,
How do you calculate Sum{k=1 to n} k.(2^k) ?

You could take a geometric series
1 + x + x^2 + ... + x^{n-1}
and differentiate it.

Yes...
1 + x + x^2 + ... + x^(n-2) = [x^n - 1] / [x - 1]

1 + 2x + 3x^2 + ... + (n-2)x^(n-3)
= [(n-1).{x^(n-2)}.(x-1) - {x^(n-1) - 1}] / [(x-1)^2]

x + 2x^2 + 3x^3 + ... + (n-2)x^(n-2)
= x*( [(n-1).{x^(n-2)}.(x-1) - {x^(n-1) - 1}] / [(x-1)^2] )

2 + 2.2^2 + 3.2^3 + ... + (n-2).2^(n-2)
= {(n-1).2^(n-1)} - 2^n + 2

Thank you very much...


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