If the coefficients of x^7 in left(ax^2+frac{ | vedclass

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

Question

(B) For the expansion $\left(ax^2+\frac{1}{2bx}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax^2)^{11-r} (\frac{1}{2bx})^r = {}^{11}C_r a^

Vedclass · Accepted Answer

$729ab = 32$

Vedclass · Answer

(B) For the expansion $\left(ax^2+\frac{1}{2bx}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax^2)^{11-r} (\frac{1}{2bx})^r = {}^{11}C_r a^{11-r} (\frac{1}{2b})^r x^{22-3r}$.Setting $22-3r = 7$,we get $3r = 15$,so $r = 5$. The coefficient is ${}^{11}C_5 a^6 (\frac{1}{2b})^5 = \frac{{}^{11}C_5 a^6}{32b^5}$.For the expansion $\left(ax-\frac{1}{3bx^2}\right)^{11}$,the general term is $T_{r+1} = {}^{11}C_r (ax)^{11-r} (-\frac{1}{3bx^2})^r = {}^{11}C_r a^{11-r} (-\frac{1}{3b})^r x^{11-3r}$.Setting $11-3r = -7$,we get $3r = 18$,so $r = 6$. The coefficient is ${}^{11}C_6 a^5 (-\frac{1}{3b})^6 = \frac{{}^{11}C_6 a^5}{729b^6}$.Equating the coefficients: $\frac{{}^{11}C_5 a^6}{32b^5} = \frac{{}^{11}C_6 a^5}{729b^6}$.Since ${}^{11}C_5 = {}^{11}C_6 = 462$,we have $\frac{a^6}{32b^5} = \frac{a^5}{729b^6}$.Dividing by $a^5$ and multiplying by $b^6$,we get $\frac{ab}{32} = \frac{1}{729}$,which implies $729ab = 32$.

If the coefficients of $x^7$ in $\left(ax^2+\frac{1}{2bx}\right)^{11}$ and $x^{-7}$ in $\left(ax-\frac{1}{3bx^2}\right)^{11}$ are equal,then

Similar Questions

The ratio of the coefficient of the middle term in the expansion of $(1+x)^{20}$ and the sum of the coefficients of two middle terms in the expansion of $(1+x)^{19}$ is $....$

Let $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$s_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Statement $-1$: $s_3 = 55 \times 2^9$
Statement $-2$: $s_1 = 90 \times 2^8$ and $s_2 = 10 \times 2^8$

The constant term in the expansion of $\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$ is

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

If the $7^{th}$ term in the binomial expansion of $\left( \frac{3}{\sqrt[3]{84}} + \sqrt{3} \ln x \right)^9, x > 0,$ is equal to $729,$ then $x$ can be

Vedclass Test Series

Exam Paper Generator

Online Exam Module