The coefficient of x - 5 in the binomial expansion of x + 1 x 2 3 - x 1 3 + 1 - x - 1 x - x 1 2 10 where x ≠ 0,1 is

x 1 3 + 1 x 2 3 - x 1 3 + 1 x 2 3 - x 1 3 + 1 - x - 1 x + 1 x x - 1 10 = x 1 3 + 1 - 1 - 1 x 1 2 10 = x 1 3 - x - 1 2 10 t r + 1 = 10 C r x 1 3 10 - r x − 1 2 r ( − 1 ) r t r + 1 = 10 C r x 10 3 - 5 r 6 - 1 r ∵ 10 3 - 5 r 6 = - 5 5 r 6 = 50 6 ⇒ r = 10 Hence, 10 C 10 - 1 10 = 1

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremGeneralHard2 minPYQ_2017

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The coefficient of $x^{- 5}$ in the binomial expansion of ${(\frac{x + 1}{x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x^{\frac{1}{2}}})}^{10}$ where $x \neq 0,1$ is

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Solution:

${[\frac{(x^{\frac{1}{3}} + 1) (x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1)}{(x^{\frac{2}{3}} - x^{\frac{1}{3}} + 1)} - \frac{(\sqrt{x} - 1) (\sqrt{x} + 1)}{\sqrt{x} (\sqrt{x} - 1)}]}^{10}$

$= {(x^{\frac{1}{3}} + 1 - 1 - \frac{1}{x^{\frac{1}{2}}})}^{10}$ $= {(x^{\frac{1}{3}} - x^{\frac{- 1}{2}})}^{10}$

$t_{r + 1} =^{10} C_{r} {(x^{\frac{1}{3}})}^{10 - r} {(x^{\frac{- 1}{2}})}^{r}$ ${(- 1)}^{r}$

$t_{r + 1} =^{10} C_{r} (x^{\frac{10}{3} - \frac{5 r}{6}}) {(- 1)}^{r}$

$∵ \frac{10}{3} - \frac{5 r}{6} = - 5$

$\frac{5 r}{6} = \frac{50}{6} \Rightarrow r = 10$

Hence, $^{10} C_{10} {(- 1)}^{10} = 1$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:General

⏱ 2mℹ️ Source: PYQ_2017

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