Author Topic: problems (Read 4016 times) Tweet Share

brendan · « **on:** March 27, 2008, 04:23:11 pm »

Derive a formula for the following sum of series:
$<br />$v + 2v^2 + 3v^3 + 4v^4 + v^5 +2v^6+3v^7+4v^8+...+v^{4k+1} + 2v^{4k+2} + 3v^{4k+3} + 4v^{4k+4} +...+ 4v^{4n}$<br />$

in which $v = (1+ i)^{-1}$ and $0<i<1$ , and $n$ is a positive integer. Simplify your answer, and express it in terms of $i, v, n$ , only.

Ahmad · « **Reply #1 on:** March 27, 2008, 07:34:50 pm »

The sum is,

$1(v + v^5 + \ldots + v^{4(n-1)+1}) +$
$2(v^2 + v^6 + \ldots + v^{4(n-1)+2}) +$
$3(v^3 + v^7 + \ldots + v^{4(n-1)+3}) +$
$4(v^4 + v^8 + \ldots + v^{4(n-1)+4})$

Which is,

$1v(1 + v^4 + \ldots + v^{4(n-1)}) +$
$2v^2(1 + v^4 + \ldots + v^{4(n-1)}) +$
$3v^3(1 + v^4 + \ldots + v^{4(n-1)}) +$
$4v^4(1 + v^4 + \ldots + v^{4(n-1)})$

$= (v+2v^2+3v^3+4v^4) \cdot \frac{v^{4n} - 1}{v^4 - 1}$

enwiabe · « **Reply #2 on:** April 03, 2008, 02:23:20 pm »

Ahmadinated.

Ahmadinate (verb) - To mathematically destroy (or, as internet lingo would have it, pwn) somebody, something or a question/problem.

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