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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}}

Vedclass · Accepted Answer

$\frac{1001}{x}$

Vedclass · Answer

(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

Similar Questions

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

The coefficient of ${x^{-9}}$ in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is

Let $n$ be a positive even integer. If the ratio of the largest coefficient and the $2^{nd}$ largest coefficient in the expansion of $(1+x)^{n}$ is $11:10$, then the number of terms in the expansion of $(1+x)^{n}$ is:

Find the $13^{\text{th}}$ term in the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x \neq 0$.

The number of integral terms in the expansion of $(5^{1/2} + 7^{1/6})^{642}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module