Question

Find the coefficient of \({x^{12}}\) in the power series of each of these functions.

a) \(1/(1 + 3x)\)

b) \(1/(1 + 3x)\)

c) \(1/{(1 + x)^8}\)

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

e) \({x^3}/{(1 + 4x)^2}\)

Short answer

The coefficient of \({x^{12}}\) in the power series of each of these functions is given below;

(a) 531,441

(b) 53,248

(c) 50,388

(d)1,526,726,656

(e) -2,621,440

Step-by-step solution

Use Extended Binomial Theorem:

a).

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

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

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

The coefficient is thus in general \({a_k} = {( - 3)^k}\)

The coefficient of \({x^{12}}\) is then the case \(k = 12\) :

\(\begin{array}{c}{a_{12}} = {( - 3)^{12}}\\ = 531,441\end{array}\)

Use Extended Binomial Theorem:

b).

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

The coefficient is thus in general

\({a_k} = \left( {\begin{array}{*{20}{c}}{1 + k}\\k\end{array}} \right){2^k}\)

The coefficient of \({x^{12}}\) is then the case\(k = 12\):

\(\begin{array}{c}{a_{12}} = \left( {\begin{array}{*{20}{c}}{1 + 12}\\{12}\end{array}} \right){2^{12}}\\ = 13 \cdot {2^{12}}\\ = 53,248\end{array}\)

Use Extended Binomial Theorem:

c).

\(\begin{array}{c}\frac{1}{{{{(1 + x)}^8}}} = \frac{1}{{{{(1 - ( - x))}^8}}}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{8 + k - 1}\\k\end{array}} \right)} {( - x)^k}\\ = \sum\limits_{k = 0}^{ + \infty } {\left( {\begin{array}{*{20}{c}}{7 + k}\\k\end{array}} \right)} {( - 1)^k}{x^k}\end{array}\)

The coefficient is thus in general;

\({a_k} = \left( {\begin{array}{*{20}{c}}{7 + k}\\k\end{array}} \right){( - 1)^k}\)

The coefficient of \({x^{12}}\) is then the case\(k = 12\):

\(\begin{array}{c}{a_{12}} = \left( {\begin{array}{*{20}{c}}{7 + 12}\\{12}\end{array}} \right){( - 1)^{12}}\\ = \left( {\begin{array}{*{20}{l}}{19}\\{12}\end{array}} \right)\\ = 50,388\end{array}\)

Use Extended Binomial Theorem:

d).

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

The coefficient is thus in general;

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

The coefficient of \({x^{12}}\) is then the case\(k = 12\):

\(\begin{array}{c}{a_{12}} = \left( {\begin{array}{*{20}{c}}{2 + 12}\\{12}\end{array}} \right){4^{12}}\\ = \frac{{14(13)}}{2} \cdot {4^{12}}\\ = 1,526,726,656\end{array}\)

Use Extended Binomial Theorem:

e).

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

The coefficient is thus in general;

\({a_{k + 3}} = \left( {\begin{array}{*{20}{c}}{1 + k}\\k\end{array}} \right){( - 4)^k}\)

The coefficient of \({x^{12}}\) is then the case\(k = 9\):

\({a_{12}} = \left( {\begin{array}{*{20}{c}}{1 + 9}\\9\end{array}} \right){( - 4)^9} = - 10 \cdot {4^9} = - 2,621,440\)