p303 q23

question:

supposing that $\displaystyle S= 1+2z+3z^2+...+nz^{n-1}$ , where z not equal to 1

by considering (1-z)S, show that $\displaystyle S = \frac {1-(n+1)z^n + nz^{n+1}}{(1-z)^2}$

thanks!

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- May 21st 2008, 11:39 AMafeasfaerw23231233complex no. question no. 7
p303 q23

question:

supposing that $\displaystyle S= 1+2z+3z^2+...+nz^{n-1}$ , where z not equal to 1

by considering (1-z)S, show that $\displaystyle S = \frac {1-(n+1)z^n + nz^{n+1}}{(1-z)^2}$

thanks! - May 21st 2008, 11:44 AMMoo
Hello :D

$\displaystyle \begin{aligned} (1-z)S &=(1-z)+2z(1-z)+3z^2(1-z)+\dots+nz^{n-1}(1-z) \\

&=1-z+2z-2z^2+3z^2-3z^3+\dots+nz^{n-1}-nz^n \\

&={\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 :)