# Thread: urgent* binomial series help(work included)

1. ## urgent* binomial series help(work included)

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

1 / (2+x)^3

the answer in the back says it's (-1)^n(n+1)(n+2) * x^n all over 2^n+4
can someone help me where i went wrong or if im doing it totally wrong

My work:
[2(1+x/2)]^-3 = (1/8)(1+x/2)^-3

(1/8)*sum (-3 n)(x/2)^n

(1/8) [ 1+(-3)(x/2) + [(-3)(-4)]/2 * (x/2)^2 +... (-3 + n - 1) ]

(1/8) * sum [(-1)^n (3*4*5....n(n+1) ]/ n!

(1/2^3) * [ (-1)^n(n+1)(n+2) ] / 2^n

=[[ (-1)^n * (n+1)(n+2) ]x^n ]/ (2^n+3) ]

thanks.

2. Originally Posted by xfyz
Use the binomial series to expand the function as a power series.

1 / (2+x)^3

the answer in the back says it's (-1)^n(n+1)(n+2) * x^n all over 2^n+4
can someone help me where i went wrong or if im doing it totally wrong

My work:
[2(1+x/2)]^-3 = (1/8)(1+x/2)^-3

(1/8)*sum (-3 n)(x/2)^n

(1/8) [ 1+(-3)(x/2) + [(-3)(-4)]/2 * (x/2)^2 +... (-3 + n - 1) ]

(1/8) * sum [(-1)^n (3*4*5....n(n+1) ]/ n!

(1/2^3) * [ (-1)^n(n+1)(n+2) ] / 2^n

=[[ (-1)^n * (n+1)(n+2) ]x^n ]/ (2^n+3) ]

thanks.
I think the only place where that calculation goes wrong is that you have lost the 2 that fails to cancel from the n!.

It's often easiest to do this type of problem by writing down the first few terms of the binomial expansion, so you can see how the pattern goes.

$[2(1+x/2)]^{-3} = {\textstyle\frac18}(1+x/2)^{-3}$

$\qquad= \frac18\left\{1 + \frac{-3}1\frac x2 + \frac{(-3)(-4)}{2!}\frac{x^2}{2^2} + \frac{(-3)(-4)(-5)}{3!}\frac{x^3}{2^3} + \frac{(-3)(-4)(-5)(-6)}{4!}\frac{x^4}{2^4} + \ldots \right\}$

$= \frac18\left\{1 - \frac31\,\frac x2 + \frac{3.4}{1.2}\,\frac{x^2}{2^2} - \frac{\rlap/3.4.5}{1.2.\rlap/3}\,\frac{x^3}{2^3} + \frac{\rlap/3.\rlap/4.5.6}{1.2.\rlap/3.\rlap/4}\,\frac{x^4}{2^4} - \ldots \right\}.$

You can see that term in x^n is going to be $(-1)^n\frac18\frac{(n+1)(n+2)}{2}\frac{x^n}{2^n} = \frac{(-1)^n(n+1)(n+2)}{2^{n+4}}x^n$.