# Binomial Series Expansion and then find an approximate value [closed]

This is from an A-Level Maths paper.

Show that $\frac{x}{(1-x)^3} = x + 3x^2 + 6x^3 + O(x^4)$ (Only first three terms of infinite series expansion are asked for) Use the result to find an approximate value of $\frac{100}{729}$.

I can do the first part without a problem. On the second part, I know substituting $x = 0.1$ is what I need to do. But how do I go about figuring out that $x = 0.1$ is indeed what I need.

Setting $\frac{x}{(1-x)^3}=\frac{100}{729}$ and solving the resulting cubic does give $x=0.1$, but I feel there must be something simpler that I'm missing.

Thanks

## closed as off-topic by Jack D'Aurizio, user91500, Claude Leibovici, R_D, user223391 Jun 24 '16 at 22:26

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack D'Aurizio, user91500, Claude Leibovici, R_D, Community
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• Your first equality cannot hold: the RHS is a polynomial, the LHS is a meromorphic function with a triple pole at $x=1$. If you are talking about Taylor expansions, please state it clearly, and have a look at meta.math.stackexchange.com/questions/5020/… – Jack D'Aurizio Jun 22 '16 at 12:01
• @NumberCruncher You have probably never heard of meromorphic functions, but just multiply across: the LHS is just $x$ the RHS is a polynomial of degree 6. You have miscopied the question. – almagest Jun 22 '16 at 12:04
• Thanks for your replies. Yes, it is an infinite series, it's meant to be LHS is approximately equal to RHS, or LHS = RHS + O(x^4) – NumberCruncher Jun 22 '16 at 13:31
• "how do I go about figuring out that $x = 0.1$ is indeed what I need." (No idea how the other comments address this, which seems to be your main, and probably only, question.) The idea is to hope that $x/(1-x)^3=100/729$ has a rational solution $x=a/b$ with $1\leqslant a<b$ relatively prime integers, or equivalently, that $3^6ab^2=10^2(b-a)^3$. Since $ab^2$ and $b-a$ are relatively prime, we know that $ab\mid 10$, and suddenly we are left with very few cases (recall that $1\leqslant a\leqslant b-1$), namely, $(a,b)=(1,2)$ or $(1,5)$ or $(1,10)$ or $(2,5)$. ... – Did Jun 23 '16 at 7:31
• ... Likewise, $3^2\mid b-a$ hence $(a,b)=(1,2)$, $(1,5)$ and $(2,5)$ are impossible. Finally, if $(a,b)=(1,10)$ works, we are done, otherwise there is no rational solution. And $(a,b)=(1,10)$ works... :-) – Did Jun 23 '16 at 7:31