# Thread: Binomial series #2 stuck!

1. ## Binomial series #2 stuck!

$f(x) = \frac{5}{(1+\frac{x}{10})^4}$

$= 5\left[1 + \frac{x}{10})^-4\right]$

$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$

But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!

2. Originally Posted by mollymcf2009
$f(x) = \frac{5}{(1+\frac{x}{10})^4}$

$= 5\left[1 + \frac{x}{10})^-4\right]$

$= 5 \left[1 + (-4)(\frac{x}{10}) + \frac{(-4)(-5)}{2!}(\frac{x}{10})^2 + \frac{(-4)(-5)(-6)}{3!}(\frac{x}{10})^3 + ... \right]$

But I don't know if this is right, or what my nth term would be, much less the power series...Thanks!!
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?

3. Originally Posted by Chris L T521
Before I start, are we dealing with a Maclaurin or Taylor Series? About what point are we constructing the series?
Sorry...

Here is the whole question:

Use the binomial series to expand the function as a power series.

$f(x) = \frac{5}{(1+\frac{x}{10})^4}$

4. Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$

for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .

So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$

$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$

_______________

Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$

Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.

So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$

etc.

5. Originally Posted by o_O
Binomial series are given by: $(1+y)^k = 1 + ky + \frac{k(k-1)}{2!}y^2$ ${\color{white}.} \ + \ \frac{k(k-1)(k-2)}{3!}y^3 + \cdots + \frac{k(k-1)(k-2)\cdots(k - n+1)}{n!}y^n + \cdots$

for any $k \in \mathbb{R}$ and if $|y| < 1$. Here, $y = \tfrac{x}{10}$ and $k = -4$ .

So: $g(x) = \left( 1 + \frac{x}{10}\right)^{-4}$

$g(x) = 1 + (-4)\left(\frac{x}{10}\right) + \frac{(-4)(-5)}{2!}\left(\frac{x}{10}\right)^2 + \frac{(-4)(-5)(-6)}{3!}\left(\frac{x}{10}\right)^3 + \cdots$ $+ \ \frac{(-4)(-5)(-6)\cdots(-4-n+1)}{n!}\left(\frac{x}{10}\right)^n + \cdots$

_______________

Let's look at the general term: $\frac{(-4)(-5)(-6)\cdots(-3-n)}{n!}\frac{x^n}{10^n}$

Just looking at the expanded series, we see that with even $n$, the coefficient is positive and with odd $n$, the coefficient is negative.

So, if we factor out the negative signs: $\frac{(-1)^{n} (4)(5)(6)\cdots(n+3)}{10^n \cdot n!} x^n \cdot {\color{red}\frac{3!}{3!}} = \frac{(-1)^{n} (n+3)(n+2)(n+1)n!}{3! \cdot 10^n \cdot n!}x^n = \cdots$

etc.
Ok, I understand everything except where the 3!/3! got there.

Thanks!!!!

6. Focus on the expression: $\frac{4 \cdot 5 \cdot 6 \cdots (n+3)}{n!}$

We can simplify it by multiplying by "1": $\frac{{\color{red} 1 \cdot 2 \cdot 3} \cdot 5 \cdot 6 \cdots (n+3)}{n! \cdot {\color{red}3!}} = \frac{(n+3)!}{n! \cdot 3!}$ $= \frac{(n+3)(n+2)(n+1)n!}{n! \cdot 3!} = \frac{(n+3)(n+2)(n+1)}{ 6}$

So you can see why I multiplied both top and bottom by $3!$ so that I can simplify that long product into a simple factorial.