If the (r + 1)^{th} term in the expansion of | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.Here,$x = \left( \frac{a}{b^{1/2}} \right)^{1

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$9$

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(A) The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.Here,$x = \left( \frac{a}{b^{1/2}} \right)^{1/3} = a^{1/3} b^{-1/6}$ and $y = \left( \frac{b}{a^{1/3}} \right)^{1/2} = b^{1/2} a^{-1/6}$.So,$T_{r+1} = {}^{21}C_r (a^{1/3} b^{-1/6})^{21-r} (b^{1/2} a^{-1/6})^r$.$T_{r+1} = {}^{21}C_r a^{(21-r)/3} b^{-(21-r)/6} b^{r/2} a^{-r/6}$.$T_{r+1} = {}^{21}C_r a^{(21-r)/3 - r/6} b^{r/2 - (21-r)/6}$.Simplifying the powers:Power of $a = \frac{42-2r-r}{6} = \frac{42-3r}{6} = 7 - \frac{r}{2}$.Power of $b = \frac{3r-21+r}{6} = \frac{4r-21}{6} = \frac{2r}{3} - \frac{7}{2}$.Given that the powers of $a$ and $b$ are equal:$7 - \frac{r}{2} = \frac{2r}{3} - \frac{7}{2}$.Multiply by $6$ to clear denominators:$42 - 3r = 4r - 21$.$63 = 7r$.$r = 9$.

If the $(r + 1)^{th}$ term in the expansion of $\left( \sqrt[3]{\frac{a}{\sqrt{b}}} + \sqrt{\frac{b}{\sqrt[3]{a}}} \right)^{21}$ has the same power of $a$ and $b$,then the value of $r$ is

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