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Expert-verified Found in: Page 549 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Find the coefficient of $${x^{10}}$$ in the power series of each of these functions.a) $${\left( {1 + {x^5} + {x^{10}} + {x^{15}} + \cdots } \right)^3}$$b) $${\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7} + \cdots } \right)^3}$$c) $$\left( {{x^4} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7}} \right)(1 + x + \left. {{x^2} + {x^3} + {x^4} + \cdots } \right)$$d) $$\left( {{x^2} + {x^4} + {x^6} + {x^8} + \cdots } \right)\left( {{x^3} + {x^6} + {x^9} + } \right. \cdots \left( {{x^4} + {x^8} + {x^{12}} + \cdots } \right)$$e) $$\left( {1 + {x^2} + {x^4} + {x^6} + {x^8} + \cdots } \right)\left( {1 + {x^4} + {x^8} + {x^{12}} + } \right. \cdots )\left( {1 + {x^6} + {x^{12}} + {x^{18}} + \cdots } \right)$$

The coefficient of $${x^{10}}$$ in the power series of each of these functions is given below;

(a) 6

(b) 3

(c) 9

(d) 0

(e) 5

See the step by step solution

## a).Step 1: UseExtended Binomial Theorem:

The Extended Binomial Theorem: Let $$x$$ be a real number with $$|x| < 1$$ and let $$u$$ be a real number. Then

$${(1 + x)^u} = \sum\limits_{k = 0}^\infty {\left( {\begin{array}{*{20}{l}}u\\k\end{array}} \right)} {x^k}$$

$$\begin{array}{c}{\left( {1 + {x^5} + {x^{10}} + {x^{15}} + \ldots } \right)^3} = {\left( {{x^{5(0)}} + {x^{5(1)}} + {x^{5(2)}} + {x^{5(3)}} + \ldots } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^{5k}}} } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{{\left( {{x^5}} \right)}^k}} } \right)^3}\\ = {\left( {\frac{1}{{1 - {x^5}}}} \right)^3}\\ = {\left( {1 + \left( { - {x^5}} \right)} \right)^{ - 3}}\\ = \sum\limits_{k = 0}^{ + \infty } {( - 3)} {\left( { - {x^5}} \right)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{5k}}\end{array}$$

## Step 2: The coefficient of $${x^{10}}$$ is then the case$$k = 2$$:

$$\begin{array}{c}{a_{5k}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\2\end{array}} \right){( - 1)^2}\\ = \frac{{( - 3)( - 4)}}{{2!}}\\ = \frac{{12}}{2}\\ = 6\end{array}$$

Thus, the coefficient of $${x^{10}}$$ is 6.

b).

## Step 3: UseExtended Binomial Theorem:

$$\begin{array}{c}{\left( {{x^3} + {x^4} + {x^5} + \ldots } \right)^3} = {\left( {{x^3}} \right)^3}{\left( {{x^0} + {x^1} + {x^2} + {x^3} + \ldots } \right)^3}\\ = {x^9} \cdot {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^k}} } \right)^3}\\ = {x^9} \cdot {\left( {\frac{1}{{1 - x}}} \right)^3}\\ = {x^9} \cdot {(1 + ( - x))^{ - 3}}\\ = {x^9} \cdot \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - x)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 9}}\end{array}$$

## Step 4: The coefficient of $${x^{10}}$$ is then the case $$k = 1$$ :

$$\begin{array}{c}{a_{k + 9}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\1\end{array}} \right){( - 1)^1}\\ = - \frac{{( - 3)}}{{1!}}\\ = 3\end{array}$$

Thus, the coefficient of $${x^{10}}$$ is 3

c).

## Step 5: UseExtended Binomial Theorem:

$$\begin{array}{c}\left( {{x^4} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + \ldots } \right)\\ = {x^4}\left( {1 + x + {x^2}} \right)\left( {{x^3}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4}} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + \ldots } \right)\\ = {x^7} \cdot \sum\limits_{k = 0}^2 {{x^k}} \cdot \sum\limits_{k = 0}^4 {{x^k}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^k}} \\ = {x^7} \cdot \sum\limits_{k = 0}^2 {{x^k}} \cdot \sum\limits_{k = 0}^4 {{x^k}} \cdot \frac{1}{{1 - x}}\\ = {x^7} \cdot \frac{{1 - {x^3}}}{{1 - x}} \cdot \frac{{1 - {x^5}}}{{1 - x}} \cdot \frac{1}{{1 - x}}\end{array}$$

By further simplification

$$\begin{array}{c} = \frac{{{x^7}\left( {1 - {x^3}} \right)\left( {1 - {x^5}} \right)}}{{{{(1 - x)}^3}}}\\ = \frac{{{x^7}\left( {1 - {x^3} - {x^5} + {x^8}} \right)}}{{{{(1 - x)}^3}}}\\ = \left( {{x^7} - {x^{10}} - {x^{12}} + {x^{15}}} \right) \cdot {(1 + ( - x))^{ - 3}}\\ = \left( {{x^7} - {x^{10}} - {x^{12}} + {x^{15}}} \right) \cdot \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - x)^k}\end{array}$$

By further simplification

$$\begin{array}{c} = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 7}} - \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 10}}\\ - \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 12}} + \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 15}}\end{array}$$

## Step 6: The coefficient of $${x^{10}}$$ is then the case $$k = 3$$ in the first summation and the case $$k = 0$$ in the second summation:

$$\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\3\end{array}} \right){( - 1)^3} - \left( {\begin{array}{*{20}{c}}{ - 3}\\0\end{array}} \right){( - 1)^0}\\ = - \frac{{( - 3)( - 4)( - 5)}}{{3!}} - 1\\ = 10 - 1\\ = 9\end{array}$$

d).

## Step 7: Use Extended Binomial Theorem:

$$\begin{array}{c}\left( {{x^2} + {x^4} + {x^6} + \ldots } \right)\left( {{x^3} + {x^6} + {x^9} + \ldots } \right)\left( {{x^4} + {x^8} + {x^{12}} + \ldots } \right)\\ = {x^2}\left( {1 + {x^2} + {x^4} + \ldots } \right)\left( {{x^3}} \right)\left( {1 + {x^3} + {x^6} + \ldots } \right)\left( {{x^4}} \right)\left( {1 + {x^4} + {x^8} + \ldots } \right)\\ = {x^9} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{2k}}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{3k}}} \cdot \sum\limits_{k = 0}^{ + \infty } {{x^{4k}}} \end{array}$$

Since the first factor is$${x^9}$$, we require a factor $$x$$ in at least one of the sums. We then note that we cannot obtain the factor $$x$$ from any of the sums (as the sums only provide the second, third, and fourth powers of $$x$$ )

Thus $${x^{10}}$$ is not present in the series and thus the coefficient of $${x^{10}}$$ has to be zero.

e).

## Step 8: Use Extended Binomial Theorem:

$$\begin{array}{l}\left( {1 + {x^2} + {x^4} + \ldots } \right)\left( {1 + {x^4} + {x^8} + \ldots } \right)\left( {1 + {x^6} + {x^{12}} + \ldots } \right)\\ = \sum\limits_{k = 0}^{ + \infty } {{x^{2k}}} \cdot \sum\limits_{m = 0}^{ + \infty } {{x^{4m}}} \cdot \sum\limits_{n = 0}^{ + \infty } {{x^{6n}}} \end{array}$$

We then obtain $${x^{10}}$$ if$$2k + 4m + 6n = 10$$. Since$$k$$, $$m$$ and $$n$$ are non-negative integers:

$$\begin{array}{*{20}{r}}{k = 5,m = 0,n = 0}\\{{\rm{ or }}k = 3,m = 1,n = 0}\\{{\rm{ or }}k = 1,m = 2,n = 0}\\{{\rm{ or }}k = 0,m = 1,n = 1}\\{{\rm{ or }}k = 2,m = 0,n = 1}\end{array}$$

Since the coefficient of each combination is 1 and since there are 5 combinations, the coefficient of $${x^{10}}$$ is:

$$1 + 1 + 1 + 1 + 1 = 5$$ ### Want to see more solutions like these? 