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Expert-verified Found in: Page 551 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Find the sequence with each of these functions as its exponential generating function${\mathbit{f}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbit{e}}}^{\mathbf{3}\mathbf{x}}{\mathbf{-}}{\mathbf{3}}{{\mathbit{e}}}^{\mathbf{2}\mathbf{x}}$.

${a}_{k}={3}^{k}-3·{2}^{k}$.

See the step by step solution

## Step 1: Given data

The given function is $f\left(x\right)={e}^{3x}-3{e}^{2x}$

## Step 2: Concept used of generating function

The ordinary generating function of a sequence ${{\mathbit{a}}}_{{\mathbf{n}}}$ is

${\mathbit{G}}\mathbf{\left(}{\mathbf{a}}_{\mathbf{n}}\mathbf{;}\mathbf{x}\mathbf{\right)}{\mathbf{=}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{{\mathbit{a}}}_{{\mathbf{n}}}{{\mathbit{x}}}^{{\mathbf{n}}}{\mathbf{.}}$

## Step 3: Solve the function

Using $\sum _{k=0}^{+\infty }\frac{{a}_{k}}{k!}{x}^{k}={e}^{x}$, we have

${e}^{3x}-3{e}^{2x}$=$\sum _{k=0}^{+\infty }\frac{{\left(3x\right)}^{k}}{k!}-3\sum _{k=0}^{+\infty }\frac{{\left(2x\right)}^{k}}{k!}\phantom{\rule{0ex}{0ex}}$

$=\sum _{k=0}^{+\infty }{3}^{k}\frac{{x}^{k}}{k!}-3\sum _{k=0}^{+\infty }{2}^{k}\frac{{x}^{k}}{k!}$

$=\sum _{k=0}^{+\infty }\left({3}^{k}-3·{2}^{k}\right)\frac{{x}^{k}}{k!}$

The sequence $\left\{{a}_{k}\right\}$ are then the coefficients of $\frac{{x}^{k}}{k!}$in the above sum: ${a}_{k}={3}^{k}-3·{2}^{k}$. ### Want to see more solutions like these? 