What is he term independent of $x$ in the expansion of $$ \Bigg[\left(\dfrac{x+1 }{ x^{2/3} - x^{1/3} + 1}\right ) - \left(\dfrac{x - 1 }{ x-x^{1/2}}\right)\Bigg]^{10} $$
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You do know the binomial theorem, yes? – J. M. ain't a mathematician Apr 08 '13 at 15:31
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Yea but i'm having problem in dealing with x^2/3 and x^1/3 – user71589 Apr 08 '13 at 15:32
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Note that 1/x^2/3 should be read 1/(x^2)/3 which is presumably $\frac 1{3x^2}$ (or perhaps $\frac 3{x^2}$)and not what you meant. Please use parentheses, or, preferably $\LaTeX$. You can get some help here – Ross Millikan Apr 08 '13 at 15:34
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I wrote the general term T(r+1)=nCr(x+1/x^2/3-x^1/3+1)^10-r-1)^r(x-1/x-x^1/2)^r – user71589 Apr 08 '13 at 15:34
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Its actually x^(2/3).I for got to use parenthesis. – user71589 Apr 08 '13 at 15:38
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I'm having trouble figuring out what that expression's meant to be. Try using these to make your expression clearer: 1) Enclose the expression in dollar ($) signs to format it, 2) \frac{a}{b} gives $\frac{a}{b}$, 3) e^{2x} gives $e^{2x}$ – Sp3000 Apr 08 '13 at 15:47
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Please check that my edit didn't do undue violence to your intent, my answer and @exploringnet's start with very different interpretations. – vonbrand Apr 08 '13 at 16:27
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@user71589 I hope you have got what you want. – ABC Apr 08 '13 at 16:37
2 Answers
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A much simpler proof: Try to simplify.
$$ \begin{align*} \left( \dfrac{x + 1}{x^{2/3} - x^{1/3} + 1} - \dfrac{x - 1}{x - x^{1/2}} \right)^{10} &= \left( (x^{1/3} + 1) - \dfrac{(\sqrt{x} - 1) (\sqrt{x} + 1)}{\sqrt{x} ( \sqrt{x} - 1)} \right)^{10} \\ &= (x^{1/3} + 1 - 1 - x^{-1/2})^{10} \\ &= (x^{1/3} - x^{-1/2})^{10} \end{align*} $$ The general term is: $$ \binom{10}{r} x^{r/3} x^{-(10 - r)/2} $$ So we want $r$ such that: $$ \frac{r}{3} - \frac{10 - r}{2} = 0 $$ This gives $r = 6$, and the coefficient is $$\binom{10}{6} = \binom{10}{4} = 210$$
ABC
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1Please don't add images when LaTeX will do. Please don't use that bletcherous notation for binomial coefficients, the civilized world has long agreed they are written
\binom{n}{k}in LaTeX, rendered $\binom{n}{k}$. – vonbrand Apr 08 '13 at 16:25 -
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The roots can be eliminated by the substitution $x \mapsto u^6$. The development will be essentially the same, just less root signs to make you pause... – vonbrand Apr 08 '13 at 16:47
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@vonbrand Well in exam this substitution was first method in my brain but afterwards i simplified it. – ABC Apr 08 '13 at 16:50
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To simplify the mess, substitute $x \mapsto u^6$, and multiply the parentesis through by $u^6$: $$ \begin{align*} \left( (u^6 + u^{-4} - u^{-2} + 1) - (u^6 - u^{-6} - u^{-3}) \right)^{10} &= u^{-60} (u^{12} + u^2 - u^4 + u^6 - u^{12} + 1 + u^3)^{10} \\ &= u^{-60} (u^6 - u^4 + u^3 + u^2 + 1)^{10} \end{align*} $$ The only way to get $u^{60}$ is by $(u^6)^{10}$; by the binomial theorem its coefficient is 1.
vonbrand
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You interpreted it wrong. Well you can solve it by this different approach again. – ABC Apr 08 '13 at 16:33
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