complex no. question no. 7

**afeasfaerw23231233** · May 21st 2008, 11:39 AM

p303 q23
question:
supposing that $S= 1+2z+3z^2+...+nz^{n-1}$ , where z not equal to 1
by considering (1-z)S, show that $S = \frac {1-(n+1)z^n + nz^{n+1}}{(1-z)^2}$
thanks!

**Moo** · May 21st 2008, 11:44 AM

Hello

Originally Posted by afeasfaerw23231233

p303 q23
question:
supposing that $S= 1+2z+3z^2+...+nz^{n-1}$ , where z not equal to 1
by considering (1-z)S, show that $S = \frac {1-(n+1)z^n + nz^{n+1}}{(1-z)^2}$
thanks!

$\begin{aligned} (1-z)S &=(1-z)+2z(1-z)+3z^2(1-z)+\dots+nz^{n-1}(1-z) \\<br /> &=1-z+2z-2z^2+3z^2-3z^3+\dots+nz^{n-1}-nz^n \\<br /> &={\color{red}1+z+z^2+\dots+z^{n-1}}-nz^n \end{aligned}$

Note that the red part is the sum of the terms of a geometric sequence, of constant progression z.

I think you can go to the result from here, just ask if you can't

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