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Math Help - complex no. question no. 7

  1. #1
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    complex no. question no. 7

    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!
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    Quote Originally Posted by afeasfaerw23231233 View Post
    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.

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