The 11^{th} term in the expansion of left(x+f | vedclass

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.Here,$n=14$,$a=x$,$b=\frac{1}{\sqrt{x}}

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$\frac{1001}{x}$

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.Here,$n=14$,$a=x$,$b=\frac{1}{\sqrt{x}}$,and we need the $11^{th}$ term,so $r+1=11$,which implies $r=10$.Substituting these values:$T_{11} = {}^{14}C_{10} (x)^{14-10} \left(\frac{1}{\sqrt{x}}ight)^{10}$$T_{11} = {}^{14}C_{10} (x)^4 \left(\frac{1}{x^{1/2}}ight)^{10}$$T_{11} = {}^{14}C_{10} (x)^4 \left(\frac{1}{x^5}ight)$$T_{11} = {}^{14}C_{10} \cdot \frac{1}{x}$Calculating ${}^{14}C_{10} = {}^{14}C_{4} = \frac{14 	imes 13 	imes 12 	imes 11}{4 	imes 3 	imes 2 	imes 1} = 1001$.Thus,$T_{11} = \frac{1001}{x}$.

The $11^{th}$ term in the expansion of $\left(x+\frac{1}{\sqrt{x}}\right)^{14}$ is

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