Finding the coefficient of $x^r$ in an expansion.

Question

Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.

Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.

Hence show that $(217)^\frac{1}{3} \simeq 6.0092$

My attempt :

$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$

$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$

$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$

$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$

$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$

How can I proceed after this ? Is there another method ? Is my method correct ?

Answer 1

In another approach, as inquired in the OP, we write the term of interest as

$$\begin{align} \frac{1+x+x^2}{(1-x)^2}&=1-\frac{3}{1-x}+\frac{3}{(1-x)^2}\\\\ &=1-3\sum_{n=0}^\infty x^n+3\frac{d}{dx}\sum_{n=0}^\infty x^n\\\\ &=1-3\sum_{n=0}^\infty x^n+3\sum_{n=1}^\infty nx^{n-1}\\\\ &=1-3\sum_{n=0}^\infty x^n+3\sum_{n=0}^\infty (n+1)x^n\\\\ &=1+3\sum_{n=0}^\infty nx^n \end{align}$$

Hence, the coefficient on the $n$'th term is indeed $3n$ as was to be shown!

Answer 2

If we MUST use the binomial coefficients, this becomes rather tedious: $$\frac{1+x^2+x}{(1-x)^2}=\frac{(1-x)^2+3x}{(1-x)^2}=1+\frac {3x}{(1-x)^2}$$$$=1-3x(1-x)^{-2}=1-3(1-x)^{-1}+3(1-x)^{-2}$$ $$=1-3\sum\binom{-1}{r}(-1)^rx^r+3\sum\binom{-2}{r}(-1)^rx^r$$ $$=1+3\sum \left(-\binom{-1}{r}+\binom{-2}{r}\right)(-1)^rx^r$$

Let's try to prove that the binomial difference if equal to $(-1)^r\times r$. For $r=0$ the sum is $0$, for $k=1$ the sum is $-1$, assuming this is true for $r=m$, let's check for $r=m+1$: $$-\binom{-1}{m+1}+\binom{-2}{m+1}=-\frac{-1-m}{1+m}\binom{-1}{m}+\frac{-2-m}{1+m}\binom{-2}{m}$$ $$=(-1)m(-1)^m-\frac 1{m+1}\binom{-2}{m}=m(-1)^{m+1}+(-1)^{m+1}$$ So the expansion becomes $$1+\sum_r3rx^r$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & = {1 - x^{3} \over \pars{1 - x}^{3}} = \pars{1 - x^{3}}\sum_{r = 0}^{\infty}{-3 \choose r}\pars{-x}^{r} = \half\pars{1 - x^{3}}\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} \end{align}

Note that $\ds{{-3 \choose r} = {-\pars{-3} + r - 1 \choose r}\pars{-1}^{r} = {r + 2 \choose r}\pars{-1}^{r} = \half\pars{r + 2}\pars{r + 1}\pars{-1}^{r}}$

\begin{align} \color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & = \half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} - \half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r + 3} \\[5mm] & = \half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} - \half\sum_{r = 3}^{\infty}\pars{r - 1}\pars{r - 2}x^{r} \\[5mm] & = 1 + 3x + 6x^{2} + \half\sum_{r = 3}^{\infty}\bracks{% \pars{r + 2}\pars{r + 1} - \pars{r - 1}\pars{r - 2}}x^{r} \\[5mm] & = 1 + 3x + 6x^{2} + \sum_{r = 3}^{\infty}\pars{3r}x^{r} = 1 + \sum_{r = 1}^{\infty}\pars{3r}x^{r} \end{align}

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$$ \color{#f00}{\bracks{x^{r}}\bracks{{1 + x + x^{2} \over \pars{1 - x}^{2}}}} = \color{#f00}{\left\lbrace\begin{array}{rcrcl} \ds{1} & \mbox{if} & \ds{r} & \ds{=} & \ds{0} \\[2mm] \ds{3r} & \mbox{if} & \ds{r} & \ds{=} & \ds{1,2,3,\ldots} \\[2mm] \ds{0} &&&& \mbox{otherwise} \end{array}\right.} $$