Question

Find the coefficient of \({x^9}\) in the power series of each of these functions.

a) \({\left( {1 + {x^3} + {x^6} + {x^9} + \cdots } \right)^3}\)

b) \({\left( {{x^2} + {x^3} + {x^4} + {x^5} + {x^6} + \cdots } \right)^3}\)

c) \(\left( {{x^3} + {x^5} + {x^6}} \right)\left( {{x^3} + {x^4}} \right)\left( {x + {x^2} + {x^3} + {x^4} + \cdots } \right)\)

d) \(\left( {x + {x^4} + {x^7} + {x^{10}} + \cdots } \right)\left( {{x^2} + {x^4} + {x^6} + {x^8} + } \right.\)\( \cdots )\)

e) \({\left( {1 + x + {x^2}} \right)^3}\)

Short answer

The coefficient of \({x^9}\) in the power series of each of the given functions is given below;

(a) 10

(b) 10

(c) 3

(d) 2

(e) 0

Step-by-step solution

a). Step 1: Use Extended Binomial Theorem:

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

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

\(\begin{array}{c}{\left( {1 + {x^3} + {x^6} + {x^9} + \ldots } \right)^3} = {\left( {{x^{3(0)}} + {x^{3(1)}} + {x^{3(2)}} + {x^{3(3)}} + \ldots } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^{3k}}} } \right)^3}\\ = {\left( {\sum\limits_{k = 0}^{ + \infty } {{{\left( {{x^3}} \right)}^k}} } \right)^3}\\ = {\left( {\frac{1}{{1 - {x^3}}}} \right)^3}\\ = {\left( {1 + \left( { - {x^3}} \right)} \right)^{ - 3}}\\ = \sum\limits_{k = 0}^{ + \infty } {( - 3)} {\left( { - {x^3}} \right)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {( - 3)} {( - 1)^k}{x^{3k}}\end{array}\)

The coefficient of \({x^9}\) is then the case \(k = 3\) :

\(\begin{array}{c}{a_{3k}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_9} = \left( {\begin{array}{*{20}{c}}{ - 3}\\3\end{array}} \right){( - 1)^3}\\ = - \frac{{( - 3)( - 4)( - 5)}}{{3!}}\\ = 10\end{array}\)

b).

Use Extended Binomial Theorem:

\(\begin{array}{c}{\left( {{x^2} + {x^3} + {x^4} + \ldots } \right)^3} = {\left( {{x^2}} \right)^3}{\left( {{x^0} + {x^1} + {x^2} + {x^3} + \ldots } \right)^3}\\ = {x^6} \cdot {\left( {\sum\limits_{k = 0}^{ + \infty } {{x^k}} } \right)^3}\\ = {x^6} \cdot {\left( {\frac{1}{{1 - x}}} \right)^3}\\ = {x^6} \cdot {(1 + ( - x))^{ - 3}}\\ = {x^6} \cdot \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - x)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right)} {( - 1)^k}{x^{k + 6}}\end{array}\)

The coefficient of \({x^9}\) is then the case \(k = 3\) :

\(\begin{array}{c}{a_{k + 6}} = \left( {\begin{array}{*{20}{c}}{ - 3}\\k\end{array}} \right){( - 1)^k}\\{a_9} = \left( {\begin{array}{*{20}{c}}{ - 3}\\3\end{array}} \right){( - 1)^3}\\ = - \frac{{( - 3)( - 4)( - 5)}}{{3!}}\\ = 10\end{array}\)

c).

Use Extended Binomial Theorem:

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

The coefficient of \({x^9}\) is then the sum of the coefficients \({a_m}{b_k}\) that:

\({x^9}\)can then be obtained if\(7 + m + k = 9\). Since \(m\) and \(k\) are nonnegative integers.

\(\begin{array}{*{20}{r}}{m = 0,k = 2}\\{{\rm{ or }}m = 1,k = 1}\\{{\rm{ or }}m = 2,k = 0}\end{array}\)

The coefficient of \({x^9}\) is;

\(\begin{array}{c}{a_9} = {a_0}{b_2} + {a_1}{b_1} + {a_2}{b_0}\\ = 1 + 1 + 1\\ = 3\end{array}\)

d).

Use Extended Binomial Theorem:

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

\({x^9}\)can then be obtained if\(3 + 3m + 2k = 9\). Since \(m\) and \(k\) are nonnegative integers.

\(\begin{array}{*{20}{r}}{m = 2,k = 0}\\{{\rm{ or }}m = 0,k = 3}\end{array}\)

Since the coefficient of each combination is 1 and since there are 5 combinations, the coefficient of \({x^9}\) is:

\(1 + 1 = 2\)

e).

The coefficient of \({x^9}\)in the function:

\(\begin{array}{c}{\left( {1 + x + {x^2}} \right)^3} = \left( {1 + x + {x^2}} \right)\left( {1 + x + {x^2} + x + {x^2} + {x^3} + {x^2} + {x^3} + {x^4}} \right)\\ = \left( {1 + x + {x^2}} \right)\left( {1 + 2x + 3{x^2} + 2{x^3} + {x^4}} \right)\\ = 1 + 2x + 3{x^2} + 2{x^3} + {x^4} + x + 2{x^2} + 3{x^3} + 2{x^4} + {x^5} + {x^2} + 2{x^3} + 3{x^4} + 2{x^5} + {x^6}\\ = {x^6} + 3{x^5} + 6{x^4} + 7{x^3} + 6{x^2} + 3x + 1\end{array}\)

We then note that the power series does not contain \({x^9}\) and thus the coefficient of \({x^9}\) having to be zero.

\({a_9} = 0\)