# Binomial Series

(a) Expand

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

as a power series.

(b) Use part (a) to find the sum of the series

$$\sum _{n=0} ^{\infty} \frac{n^2}{2^n}$$

$$\hline$$

Here is what I've got:

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x \right) ^3 } = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{k} (-x) ^k = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^k \binom{-3}{k} x ^k$$

$$f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right]$$

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

$$f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right]$$

$$f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right]$$

(b)

$$\sum _{n=0} ^{\infty} \frac{n^2}{2^n} = ?$$

Thank you

Last edited:

vsage
Did you write the question correctly? I don't see how you can use partial fractions in the last equation you give to get the question again. I don't think you can either because for x = 2 the equivalency isn't there.

Last edited by a moderator:
arildno
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thiago_j said:
(a) Expand

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

as a power series.

(b) Use part (a) to find the sum of the series

$$\sum _{n=0} ^{\infty} \frac{n^2}{2^n}$$

$$\hline$$

Here is what I've got:

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x^3\right)} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3}$$
That last equality is incorrect!

I see. You guys are right about that last equality.

If I have

$$\sum _{n=0} ^{\infty} \frac{n^2}{2^n}$$

I may need to rewrite

$$\frac{n^2}{2^n}$$

and compute it with the aid of the f(x) I found in part (a). What I can do is:

$$\frac{n^2}{2^n} = n^2 \left( \frac{1}{2} \right) ^n$$

but the connection with f(x) still is not clear.

I've just made some corrections to my 1st post.... did I get it right this time?

Thanks.

vsage,

I'm sorry. I wrote it incorrectly last time. It isn't:

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

I've just edited my 1st post to fix it. Here it is:

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

Sorry again, and thanks for your help.

vsage
Consider that $$\frac {1}{(1-x)^3} = \frac {d^2\frac{1}{2} \frac{1}{1-x}}{dx^2}$$

Well, I see that:

$$f(x) = \frac{x+x^2}{\left( 1-x\right) ^3} = -\frac{1}{2}\left( x + x^2 \right) \frac{d^2}{dx^2} \left[ \frac{1}{1-x} \right] = -\frac{1}{2}\left( x + x^2 \right) \frac{d^2}{dx^2} \left[ \sum _{n=0} ^{\infty} x^n \right] = -\frac{1}{2}\left( x + x^2 \right) \sum _{n=2} ^{\infty} n(n-1)x^{n-2}$$

This method is, in fact, simpler. The directions of this problem in part (a) are to expand f(x) as a power series, so it's ok to do that. But, I've got this exercise from the "Binomial Series" section in my calc book, so I tried to stick to it. That's why I used it in my 1st post.

Maybe your tip was about the connection between f(x) and the series in part (b). To be honest, I don't see how it may be related to it.

Folks, let me rephrase what I have. I was able to develop the solution quite a bit farther, but there still are some unclear areas.

Problem:

(a) Expand

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

as a power series.

(b) Use part (a) to find the sum of the series

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n}$$

$$\hline$$

Solution:

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{k} (-x) ^k = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^k \binom{-3}{k} x ^k$$

$$f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right]$$

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

$$f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right]$$

$$f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right]$$

$$f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} \qquad \fbox{UNCLEAR 1}$$

$$f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right]$$

$$f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3}$$

(b)

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n} = 6 = f\left( \frac{1}{2} \right) \qquad \fbox{UNCLEAR 2}$$

$$\fbox{UNCLEAR 1}:$$ I don't understand the transition from the series above to this result.
$$\fbox{UNCLEAR 2}:$$ I've found the sum of the series ("6") with the aid of my calculator. Later, I evaluated a guess ("1/2"), which gave the correct answer. However, it still isn't explicit the connection between

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n} \qquad \mbox{ and } \qquad f(x) = \frac{x+x^2}{\left( 1-x\right) ^3}$$

Thank you very much.

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Guys, I may have found the rest of the solution:

(a)

$$f(x)=\frac{x+x^2}{\left( 1-x\right) ^3} = \left( x + x^2 \right) \left[ 1 + (-x) \right] ^{-3} = \left( x + x^2 \right) \sum _{n=0} ^{\infty} \binom{-3}{n} (-x) ^n = \left( x + x^2 \right) \sum _{n=0} ^{\infty} (-1) ^n \binom{-3}{n} x ^n$$

$$f(x) = \left( x + x^2 \right) \left[ \binom{-3}{0} - \binom{-3}{1}x + \binom{-3}{2} x^2 - \binom{-3}{3} x^3 + \binom{-3}{4} x^4 - \cdots \right]$$

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

$$f(x) = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} \frac{\left( n+2 \right)!}{n!}x^n \right] = \left( x + x^2 \right) \left[ 1 + \frac{1}{2} \sum _{n=1} ^{\infty} (n+2)(n+1)x^n \right]$$

$$f(x) = x + x^2 + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+1} \right] + \frac{1}{2} \sum _{n=1} ^{\infty} \left[ (n+2)(n+1)x^{n+2} \right]$$

$$f(x) = x + x^2 - \frac{x\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} - \frac{x^2\left( 3x-3x^2+x^3\right)}{\left( x-1 \right) ^3} \qquad \fbox{UNCLEAR}$$

$$f(x) = x + x^2 - \left[ x +\frac{1}{\left( x-1 \right) ^2} +\frac{1}{\left( x-1 \right) ^3} \right] - \left[ x^2 + \frac{1}{ x-1} + \frac{2}{\left( x-1 \right) ^2} + \frac{1}{\left( x-1 \right) ^3} \right]$$

$$f(x) = -\frac{1}{ x-1} -\frac{3}{\left( x-1 \right) ^2} - \frac{2}{\left( x-1 \right) ^3} = \frac{x\left( x+1 \right)}{\left( x-1 \right) ^3} = x \frac{d}{dx} x \frac{d}{dx} \left[ \frac{1}{1-x} \right] = x \frac{d}{dx} x \frac{d}{dx} \left[ \sum _{n=0} ^{\infty} x^n \right]$$

$$f(x) = \sum _{n=1} ^{\infty} n^2 x^n$$

(b)

$$\sum _{n=1} ^{\infty} \frac{n^2}{2^n} = \sum _{n=1} ^{\infty} n^2 \left( \frac{1}{2} \right) ^n = f\left( \frac{1}{2} \right) = \frac{\frac{1}{2}+\left( \frac{1}{2} \right)^2}{\left( 1-\frac{1}{2}\right) ^3} = 6$$

$$\fbox{UNCLEAR}:$$ I don't understand the transition from the series above to this result.