Question

(a) Expand \(f(x) = (x + {x^2})/{(1 - x)^3}\) as a power series.

(b) Use part (a) to find the sum of the series \(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} \).

Short answer

\(\frac{{x + {x^2}}}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {{n^2}{x^n}} \)

Step-by-step solution

(a) Expand \(f(x) = (x + {x^2})/{(1 - x)^3}\) as a power series.Step 1: Given series and simplification

We know that

\(\frac{1}{{{{(1 - x)}^3}}} = \frac{1}{{{{[1 + ( - x)]}^3}}} = {[1 + ( - x)]^{ - 3}}\)

Step 2: Expand

\(\begin{array}{l}\frac{1}{{{{(1 - x)}^3}}} = 1 + ( - 3)( - x) + \frac{{( - 3)( - 4)}}{{2!}}{( - x)^2} + \frac{{( - 3)( - 4)( - 5)}}{{3!}}{( - x)^3}\\ + \frac{{( - 3)( - 4)( - 5)( - 6)}}{{4!}}{( - x)^4} + ......\end{array}\)\(\frac{1}{{{{(1 - x)}^3}}} = 1 + 3x + 6{x^2} + 10{x^3} + 15{x^4} + ......\)

Step 3: Further simplification

Therefore we have

\(\begin{array}{l}\frac{{x + {x^2}}}{{{{(1 - x)}^3}}} = (x + {x^2})(1 + 3x + 6{x^2} + 10{x^3} + 15{x^4} + ......)\\\frac{{x + {x^2}}}{{{{(1 - x)}^3}}} = x + 3{x^2} + {x^2} + 6{x^3} + 3{x^3} + 10{x^4} + 16{x^4} + 15{x^5} + 10{x^5} + .......\\\frac{{x + {x^2}}}{{{{(1 - x)}^3}}} = x + 4{x^2} + 9{x^3} + 16{x^4} + 25{x^5} + .....\\\frac{{x + {x^2}}}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {{n^2}{x^n}} \end{array}\)

 (b) Use part (a) to find the sum of the series \(\sum\limits_{n = 1}^\infty  {\frac{{{n^2}}}{{{2^n}}}} \). \(\sum\limits_{n = 1}^\infty  {\frac{{{n^2}}}{{{2^n}}} = 6} \)Step 1: Given series and simplification

We have the series given as \(S = \sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} = \sum\limits_{n = 1}^\infty {{n^2}} {\left( {\frac{1}{2}} \right)^n}\)

We know that \(\frac{{x + {x^2}}}{{{{(1 - x)}^3}}} = \sum\limits_{n = 1}^\infty {{n^2}{x^n}} \)

Step 2: Substitution and simplification

Putting \(x = \frac{1}{2}\) we have :

\(\begin{array}{l}\frac{{x + {x^2}}}{{{{(1 - x)}^3}}} = \sum\limits_{n = 1}^\infty {{n^2}{x^n}} \\\sum\limits_{n = 1}^\infty {{n^2}{{\left( {\frac{1}{2}} \right)}^n}} = \left[ {\frac{{\left( {\frac{1}{2}} \right) + {{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {1 - \left( {\frac{1}{2}} \right)} \right)}^3}}}} \right]\end{array}\)\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} = \left[ {\frac{{\left( {\frac{1}{2}} \right) + \left( {\frac{1}{4}} \right)}}{{{{\left( {\frac{1}{2}} \right)}^3}}}} \right]\)\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} = \left[ {\frac{{\left( {\frac{3}{4}} \right)}}{{\left( {\frac{1}{8}} \right)}}} \right]\)\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} = \frac{{(3)(8)}}{{(4)(1)}}\)\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}} = 6\)