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The constant term in expansion of \( {\left[\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right]^{10} } \) is |
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Answer» Generm term in the expansion of \([\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}]^{10}\) is Tr+1 = (-1)r 10Cr (\(\frac{x+1}{x^{2/3}-x^{1/3} + 1}\))10-r \((\frac{x-1}{x-x^{1/2}})^r\) = (-1)r 10Cr (\(\frac{x+1}{x^{2/3}-x^{1/3}+1}\))10-r \(\left(\frac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}+1)}\right)^r\) = (-1)r 10Cr \((\frac{x+1}{x^{2/3}-x^{1/3}+1})^{10-r}\)\(\frac{(x^{1/2}+1)}{x^{r/2}}\) We get a constant term if r = 10 (\(\because\) if r \(\neq\) 0 then (\(\frac{x+1}{x^{2/3}-x^{1/3}+1}\))10-r always gives terms in variables) Let r = 10 then T11 = (-1)10 10C10 \(\frac{(x^{1/2}+1)^{10}}{x^5}\) ⇒ T11 = \(\frac{(x^{1/2}+1)^{10}}{x^5}\) = \(\cfrac{^{10}C_r x^{\frac{10-r}21^r}}{x^5}\) = 10Cr \(x^{\frac{10-r}2-5}\) For constant term \(\frac{10-r}2-5 = 0\) ⇒ \(\frac{10-r}2\) = 5 ⇒ r = 10 - 10 ⇒ r = 0 Then constant term = 10C0 = 1 Hence, constant term in given expansion is 1. |
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