No.
We know that
"\\sum_{n=0}^{\\infin} {z^n} =\\frac 1 {1-z}"
Find derivative of it:
"\\sum_{n=1}^{\\infin} {n*z^{n-1}} =\\frac 1 {(1-z)^2}"
Change n to n+1
"\\sum_{n=0}^{\\infin} {(n+1)*z^{n}} =\\frac 1 {(1-z)^2}"
Hence, the generating function of the sequence {1,2,3...n..} is "\\frac 1 {(1-z)^2}"
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