Let m and n be the coefficients of the sevent | vedclass

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

Question

(D) The general term in the expansion of $(a+b)^{18}$ is $T_{r+1} = {}^{18}C_r a^{18-r} b^r$. For the seventh term $(T_7)$,$r=6$: $T_7 = {}^{18}C_6 \l

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(D) The general term in the expansion of $(a+b)^{18}$ is $T_{r+1} = {}^{18}C_r a^{18-r} b^r$. For the seventh term $(T_7)$,$r=6$: $T_7 = {}^{18}C_6 \left(\frac{1}{3} x^{\frac{1}{3}}\right)^{12} \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^6 = {}^{18}C_6 \cdot \frac{1}{3^{12}} \cdot \frac{1}{2^6} \cdot x^4 \cdot x^{-4} = {}^{18}C_6 \cdot 3^{-12} \cdot 2^{-6}$. So,$m = {}^{18}C_6 \cdot 3^{-12} \cdot 2^{-6}$. For the thirteenth term $(T_{13})$,$r=12$: $T_{13} = {}^{18}C_{12} \left(\frac{1}{3} x^{\frac{1}{3}}\right)^6 \left(\frac{1}{2} x^{-\frac{2}{3}}\right)^{12} = {}^{18}C_{12} \cdot \frac{1}{3^6} \cdot \frac{1}{2^{12}} \cdot x^2 \cdot x^{-8} = {}^{18}C_{12} \cdot 3^{-6} \cdot 2^{-12} \cdot x^{-6}$. Wait,the question asks for the coefficients of the terms. Since ${}^{18}C_6 = {}^{18}C_{12}$,we have: $\frac{n}{m} = \frac{{}^{18}C_{12} \cdot 3^{-6} \cdot 2^{-12}}{{}^{18}C_6 \cdot 3^{-12} \cdot 2^{-6}} = \frac{3^{-6}}{3^{-12}} \cdot \frac{2^{-12}}{2^{-6}} = 3^6 \cdot 2^{-6} = \left(\frac{3}{2}\right)^6$. Then $\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\left(\frac{3}{2}\right)^6\right)^{\frac{1}{3}} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

Let $m$ and $n$ be the coefficients of the seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2} x^{-\frac{2}{3}}\right)^{18}$. Then $\left(\frac{n}{m}\right)^{\frac{1}{3}}$ is:

Similar Questions

If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is .... .

For some $n \neq 10$,let the coefficients of the $5^{\text{th}}$,$6^{\text{th}}$,and $7^{\text{th}}$ terms in the binomial expansion of $(1+x)^{n+4}$ be in $A.P.$ Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:

Find the coefficient of $x^{6} y^{3}$ in the expansion of $(x+2 y)^{9}$.

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module