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Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.

I don't know a lot on power series at the moment, and I was wondering how do I find the coefficients?

Thanks

person112358
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  • You can calculate derivatives to find the first several terms of $\frac{1}{(1-t)^4}$, in fact all of them. Plug in $t=x^3$. It is obvious that the $x^{20}$ term is $0$. For $x^{21}$ you want the coefficient of $t^7$. – André Nicolas Jun 05 '15 at 19:14

3 Answers3

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HINT:

Let $y=x^3$ and expand $f(y)=(1-y)^{-4}$ in a Taylor series around $y=0$.

This series will have terms for $x^{3n}$ only, which implies that the coefficient for $x^{20}$ is zero.

We note that the $n$'th coefficient is $(n+3)!/3!n!$, from which we see that the coefficient for $x^{21}$ is at $n=7$ or $\binom{10}{3}$.

Mark Viola
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Hint:

$$\sum_{n=k}^{\infty}\binom{n}{k}y^{n-k}=\left(1-y\right)^{-k-1}$$

This can be proved by induction to $k$.

The base case is of course $\sum_{n=0}^{\infty}y^{n}=\left(1-y\right)^{-1}$.

The inductionstep comes to differentiating both sides and draw conclusions.

Apply this for $y=x^3$ and $k=3$

drhab
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Recall that $$\frac1{1-s} = \sum_{n=0}^\infty s^n. $$ Differentiating we have $$\frac{\mathsf d^3}{\mathsf ds^3}\left[\frac1{1-s}\right] = \frac6{(1-s)^4}, $$ and $$\begin{align*} \frac{\mathsf d^3}{\mathsf ds^3}\left[\sum_{n=0}^\infty s^n\right] &= \sum_{n=0}^\infty \frac{\mathsf d^3}{\mathsf ds^3}[s^n]\\ &= \sum_{n=0}^\infty n(n-1)(n-2)s^{n-3}\\ &= \sum_{n=0}^\infty (n+1)(n+2)(n+3)s^n. \end{align*}$$ Hence, $$\frac1{(1-s)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)s^n. $$ Put $s=x^3$, then $$\frac1{(1-x^3)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)x^{3n}. $$ It is clear that $x^{20}=0$ and $$x^{21} = \frac16(7+1)(7+2)(7+3) = 120. $$

Math1000
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