The numerically greatest term in the expansio | vedclass

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

Question

(C) Given expansion is $(x+3y)^{13}$ with $x=\frac{1}{2}$ and $y=\frac{1}{3}$.Substituting the values,we get $(\frac{1}{2} + 3(\frac{1}{3}))^{13} = (\

Vedclass · Accepted Answer

${ }^{13}C_9 \left(\frac{1}{2}\right)^9 \left(\frac{1}{3}\right)^4$

Vedclass · Answer

(C) Given expansion is $(x+3y)^{13}$ with $x=\frac{1}{2}$ and $y=\frac{1}{3}$.Substituting the values,we get $(\frac{1}{2} + 3(\frac{1}{3}))^{13} = (\frac{1}{2} + 1)^{13} = (\frac{3}{2})^{13}$.Let $T_{r+1}$ be the $(r+1)$-th term in the expansion of $(a+b)^n$.The condition for the numerically greatest term is $\frac{T_{r+1}}{T_r} \geq 1$.Here,$T_{r+1} = {}^{13}C_r (\frac{1}{2})^{13-r} (1)^r = {}^{13}C_r (\frac{1}{2})^{13-r}$.$\frac{T_{r+1}}{T_r} = \frac{{}^{13}C_r (\frac{1}{2})^{13-r}}{{}^{13}C_{r-1} (\frac{1}{2})^{13-(r-1)}} = \frac{{}^{13}C_r}{{}^{13}C_{r-1}} 	imes 2 = \frac{13-r+1}{r} 	imes 2 = \frac{14-r}{r} 	imes 2$.Setting $\frac{28-2r}{r} \geq 1$,we get $28-2r \geq r$,so $3r \leq 28$,which means $r \leq 9.33$.Thus,$r=9$ gives the greatest term $T_{10} = {}^{13}C_9 (\frac{1}{2})^{13-9} (1)^9 = {}^{13}C_9 (\frac{1}{2})^4$.

The numerically greatest term in the expansion of $(x+3y)^{13}$,when $x=\frac{1}{2}$ and $y=\frac{1}{3}$ is

Similar Questions

If the fourth term in the expansion of $(x+x^{\log _{2} x})^{7}$ is $4480$,then the value of $x$ where $x \in N$ is equal to

The term independent of $x$ in the expansion of $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$ is

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

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

The coefficient of $t^4$ in the expansion of ${\left( {\frac{{1 - {t^6}}}{{1 - t}}} \right)^3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module