Изменения

<tex>G(z)=\sum_sum\limits_{n=0}^\infty a_n z^n</tex>,

<tex>G(z)=a_0+a_1z+\sum_sum\limits_{n=2}^\infty(6a_{ n - 1}-8a_{n-2}+n) z^n</tex>

<tex>G(z)=a_0+a_1z+6\sum_sum\limits_{n=2}^\infty a_ { n-1}z^n - 8\sum_sum\limits_{n=2}^\infty a_ { n-2}z^n+\sum_sum\limits_{n=2}^\infty n z^n</tex>

<tex>G(z)=a_0+a_1z+6z\sum_sum\limits_{n=1}^\infty a_ { n }z^n - 8z^2\sum_sum\limits_{n=0}^\infty a_ { n }z^n+\sum_sum\limits_{n=2}^\infty n z^n</tex>

<tex>G(z)=a_0+a_1z+6z(G(z)-a_0) - 8z^2G(z)+\sum_sum\limits_{n=2}^\infty n z^n</tex>

<tex>G(z)=1-4z+6zG(z) - 8z^2G(z)+\sum_sum\limits_{n=2}^\infty n z^n</tex>

<tex>zB'(z)=z(\sum_sum\limits_{n=0}^\infty b_n z^n)'=z\sum_sum\limits_{ n = 1}^\infty nb_n z^{n-1}=\sum_sum\limits_{n=0}^\infty nb_n z^n</tex>

<tex>\sum_sum\limits_{n=2}^\infty n z^n=z \sum_sum\limits_{n=2}^\infty n z^{n-1}= z(\sum_sum\limits_{ n = 2}^\infty z^n)'</tex>

<tex>\sum_sum\limits_{n=2}^\infty z^n=\sum_sum\limits_{n=0}^\infty z^n-1-z=\dfrac{1}{1-z}-1-z=\dfrac{z^2}{1-z}</tex>

<tex>\dfrac{ 1}{(1-z)^2}=(1-z)^{-2}=\sum_sum\limits_{n=0}^{\infty} {-2\choose n}(-z)^n=</tex>

<tex>=\sum_sum\limits_{n=0}^{\infty} (-1)^n{n+1\choose 1}(-z)^n=\sum_sum\limits_{n=0}^{\infty}(n+1)z^n</tex>

<tex>=\dfrac{1}{3}\sum_sum\limits_{n=0}^{\infty} (n+1)z^n +\dfrac{7}{9}\sum_sum\limits_{n=0}^{\infty} z^n - \dfrac{1}{2}\sum_sum\limits_{n=0}^{\infty} 2^n z^n + \dfrac{7}{18}\sum_sum\limits_{n=0}^{\infty} 4^n z^n</tex>

* <tex>\operatorname{E}(\xi)=\sum_sum\limits_{n=1}^{\infty}n p(1-p)^{n-1} </tex>

* <tex>\operatorname{E}(\xi^2) = \sum_sum\limits_{n=1}^{\infty}n^{2}p(1-p)^{n-1}</tex>

* <tex>\operatorname{ E}(\xi)=\sum_sum\limits_{n=1}^{\infty}n p(1-p)^{n-1} = </tex>

<tex>= \sum_sum\limits_{n=0}^{\infty}(n+1) p(1-p)^{n} = </tex>

<tex>= \sum_sum\limits_{n=0}^{\infty}n p(1-p)^{n} + \sum_sum\limits_{n=1}^{\infty} p(1-p)^{n-1} = </tex>

* <tex>\operatorname{E}(\xi^2) = p\sum_sum\limits_{n=1}^{\infty}n^{2}(1-p)^{n-1} =</tex>

<tex>=p\sum_sum\limits_{n=1}^{\infty}n(n+1)(1-p)^{n-1} - p\sum_sum\limits_{n=1}^{\infty}n(1-p)^{n-1} =</tex>

<tex>= p\dfrac{\operatorname{d}^{2}}{\operatorname{d}p^{2}}\sum_sum\limits_{n=1}^{\infty}(1-p)^{n+1} + p\dfrac{\operatorname{d}}{\operatorname{d}p}\sum_sum\limits_{n=1}^{\infty}(1-p)^{n} =</tex>

<tex>= p\dfrac{\operatorname{d}^{2}}{\operatorname{d}p^{2}}\left(\sum_sum\limits_{ n = 0}^{\infty}(1-p)^{n} \cdot(1-p)^2\right) +p\dfrac{\operatorname{d}}{\operatorname{d}p}\left(\sum_sum\limits_{ n = 0}^{\infty}(1-p)^{n}\cdot(1-p)\right) =</tex>

| <tex>(1, 1, 1,\ldots)</tex> || <tex>\sum \limits z^n</tex> || <tex>\dfrac{1}{1-z}</tex>

| <tex>(1, 0, 0, \ldots, 0, 1, 0, 0, \ldots 0, 1, 0, 0\ldots)</tex> (повторяется через <tex>m</tex>) || <tex>\sum \limits z^{nm}</tex> || <tex>\dfrac{1}{1-z^m}</tex>

| <tex>(1, -1, 1, -1,\ldots)</tex> || <tex>\sum \limits (-1)^nz^n</tex> || <tex>\dfrac{1}{1+z}</tex>

| <tex>(1, 2, 3, 4,\ldots)</tex> || <tex>\sum \limits (n+1)z^n</tex> || <tex>\dfrac{1}{(1-z)^2}</tex>

| <tex>(1, 2, 4, 8, 16,\ldots)</tex> || <tex>\sum \limits 2^nz^n</tex> || <tex>\dfrac{1}{(1-2z)^2}</tex>

| <tex>(1, r, r^2, r^3,\ldots)</tex> || <tex>\sum \limits r^nz^n</tex> || <tex>\dfrac{1}{(1-rz)^2}</tex>

| <tex>(</tex><tex>{m\choose 0}, {m\choose 1}, {m\choose 2}, {m\choose 3},\ldots</tex><tex>)</tex> || <tex>\sum \limits {m\choose n}</tex> <tex>z^n</tex> || <tex>(1+z)^m</tex>

| <tex>(</tex><tex>1, {{m}\choose m}, {{m+1}\choose m}, {{m+2}\choose m},\ldots</tex><tex>)</tex> || <tex>\sum \limits {{m+n-1}\choose n}</tex> <tex>z^n</tex> || <tex>\dfrac{1}{(1-z)^m}</tex>

| <tex>(</tex><tex>1, {{m+1}\choose m}, {{m+2}\choose m}, {{m+3}\choose m},\ldots</tex><tex>)</tex> || <tex>\sum \limits {{m+n}\choose n}</tex> <tex>z^n</tex> || <tex>\dfrac{1}{(1-z)^{m+1}}</tex>

| <tex>(0, 1, -\dfrac{1}{2}, \dfrac{1}{3}, -\dfrac{1}{4},\ldots)</tex> || <tex>\sum \limits \dfrac{(-1)^{n+1}}{n}</tex> <tex>z^n</tex> || <tex>\ln(1+z)</tex>

| <tex>(1, 1, \dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{24},\ldots)</tex> || <tex>\sum \limits \dfrac{1}{n!}</tex> <tex>z^n</tex> || <tex>e^z</tex>