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Use binomial series to expand the function $\frac{5}{(6+x)^3}$ as a power series.

I understand the process to get the following summation:

$\frac{5}{6^3}\sum_{n=0}^{\infty} {-3 \choose n} (\frac{x}{6})^n $

However, I am stuck on seeing what's going on with ${-3 \choose n}$.

From the Stewart Calculus textbook, it says that ${k \choose n} = \frac{k(k-1)(k-2)...(k-n+1)}{n!}$.

By applying that, I would get: ${-3 \choose n}=\frac {(-3)(-4)(-5)...[-(n+2)]}{n!}$.

I think the next step would be to extract out the negative, so I will get $(-1)^n$.

The summation would then be: $\frac{5}{6^3}\sum_{n=0}^{\infty} {\frac {(-1)^n(3)(4)(5)...[(n+2)]}{n!}} (\frac{x}{6})^n $

The solution to this problem is $\frac{5}{2}\sum_{n=0}^{\infty} {\frac {(-1)^n(n+1)(n+2)x^n}{6^{n+3}}}$

I am not sure what has happened to the factorial. Was it cancelled out due to the (3)(4)(5)... in the numerator? Where did (n+1) come from?

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  • $\begingroup$ $\binom{-3}{n}=(-1)^n\binom{n+2}{n}=(-1)^n\binom{n+2}{2}=(-1)^n\frac{(n+2)(n+1)}2$ $\endgroup$
    – robjohn
    Apr 4 '19 at 6:02
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I am not sure what has happened to the factorial. Was it cancelled out due to the $(3)(4)(5)$... in the numerator? Where did $(n+1)$ come from?

It was indeed cancelled out. If you notice on the top, what you have is sort of a "falling factorial" for $(n+2)$: $(n+2)$ times the previous terms, stopping, not at $1$, but some other, earlier number ($3$ in this case). So, to write it out a bit more, we have

$$(n+2)(n+1)(n)(n-1)(n-2)...(5)(4)(3)$$

on the top. Aside from the lack of $1,2$, notice that starting at the $n$ term going onward, we have $n!$ (divided by $2!=2$ of course). Thus, all of those terms are negated since we're dividing this by $n!$, though $2$ doesn't negate so that stays around on the bottom.

Of course I imagine it's now immediately clear where the $(n+1)$ comes from.


This all aside, everything else seems to have been done correctly and the reason why the solution differs from your answer is merely them playing around with the various constants and such.

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We have the following: \begin{align} \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n+2)}{n!}\left(\frac{x}{6}\right)^n &= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n)(n+1)(n+2)}{(1)(2)(3)\ldots(n)}\left(\frac{x}{6}\right)^n\\ &= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n(n+1)(n+2)}{(1)(2)}\left(\frac{x}{6}\right)^n\\ &= \frac{5}{2}\sum_{n=0}^\infty (-1)^n(n+1)(n+2)\frac{1}{6^3}\frac{x^n}{6^n}\\ &= \frac{5}{2}\sum_{n=0}^\infty \frac{(-1)^n(n+1)(n+2)x^n}{6^{n+3}}\\ \end{align}

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$$\frac {(3)(4)(5)\cdots(n+2)}{n!} = \frac {(3)(4)(5)\cdots(n)(n+1)(n+2)}{(1)(2)(3)(4)(5)\cdots(n)} = \frac {(n+1)(n+2)}{(1)(2)}$$

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