Question

Find the coefficient of \({x^{10}}\) in the power series of each of these functions.

a) \(1/(1 - 2x)\)

b) \(1/{(1 + x)^2}\)

c) \(1/{(1 - x)^3}\)

d) \(1/{(1 + 2x)^4}\)

e) \({x^4}/{(1 - 3x)^3}\)

Short answer

The coefficient of \({x^{10}}\) in the power series of each of the given functions is given below;

(a) 1024

(b) 11

(c) 66

(d) 292,864

(e) 20,412

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

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

\(\begin{array}{c}\frac{1}{{1 - 2x}} = \sum\limits_{k = 0}^{ + \infty } {{{(2x)}^k}} \\ = \sum\limits_{k = 0}^{ + \infty } {{2^k}} {x^k}\end{array}\)

The coefficient is thus in general \({a_k} = {2^k}\)

The coefficient of \({x^{10}}\) is then the case \(k = 10\) :

\(\begin{array}{c}{a_{10}} = {2^{10}}\\ = 1024\end{array}\)

b).

Use Extended Binomial Theorem:

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

The coefficient is thus in general;

\({a_k} = \left( {\begin{array}{*{20}{c}}{1 + k}\\k\end{array}} \right){( - 1)^k}\)

The coefficient of \({x^{10}}\) is then the case\(k = 10\):

\(\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{1 + 10}\\{10}\end{array}} \right){( - 1)^{10}}\\ = 11\end{array}\)

c).

Use Extended Binomial Theorem:

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

The coefficient is thus in general;

\({a_k} = \left( {\begin{array}{*{20}{c}}{2 + k}\\k\end{array}} \right)\)

The coefficient of \({x^{10}}\) is then the case\(k = 10\):

\(\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{2 + 10}\\{10}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}}{12}\\{10}\end{array}} \right)\\ = \frac{{12 \cdot 11}}{2}\\ = 66\end{array}\)

d).

Use Extended Binomial Theorem:

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

The coefficient is thus in general;

\({a_k} = \left( {\begin{array}{*{20}{c}}{3 + k}\\k\end{array}} \right){( - 2)^k}\)

The coefficient of \({x^{10}}\) is then the case\(k = 10\):

\(\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{3 + 10}\\{10}\end{array}} \right){( - 2)^{10}}\\ = \frac{{13(12)(11)}}{{3(2)}} \cdot {2^{10}}\\ = 292,864\end{array}\)

e).

Use Extended Binomial Theorem:

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

The coefficient is thus in general;

\({a_{k + 4}} = \left( {\begin{array}{*{20}{c}}{2 + k}\\k\end{array}} \right){3^k}\)

The coefficient of \({x^{10}}\) is then the case\(k = 6\):

\(\begin{array}{c}{a_{10}} = \left( {\begin{array}{*{20}{c}}{2 + 6}\\6\end{array}} \right){3^6}\\ = \frac{{8(7)}}{{2!}} \cdot {3^6}\\ = 20,412\end{array}\)