The 16^{th} term in the expansion of (sqrt{x} | vedclass

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(C) The general term in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.For the expansion of $(\sqrt{x} - \sqrt{y})^{17}$,we

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$-136xy^{15/2}$

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(C) The general term in the expansion of $(a + b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.For the expansion of $(\sqrt{x} - \sqrt{y})^{17}$,we have $n = 17$,$a = \sqrt{x}$,and $b = -\sqrt{y}$.To find the $16^{th}$ term $(T_{16})$,we set $r+1 = 16$,which gives $r = 15$.$T_{16} = {}^{17}C_{15} (\sqrt{x})^{17-15} (-\sqrt{y})^{15}$$T_{16} = {}^{17}C_2 (\sqrt{x})^2 (-\sqrt{y})^{15}$$T_{16} = \frac{17 	imes 16}{2 	imes 1} 	imes x 	imes (-y^{15/2})$$T_{16} = 136 	imes x 	imes (-y^{15/2}) = -136xy^{15/2}$.

The $16^{th}$ term in the expansion of $(\sqrt{x} - \sqrt{y})^{17}$ is

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