If the constant term in the expansion of (1+2 | vedclass

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

Question

(B) The given expression is $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$.The general term of the expansion $(\frac{3}{2}x^2-\frac{1}{3x})^9$ is given

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) The given expression is $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$.The general term of the expansion $(\frac{3}{2}x^2-\frac{1}{3x})^9$ is given by $T_{r+1} = {}^9C_r (\frac{3}{2}x^2)^{9-r} (-\frac{1}{3x})^r$.$T_{r+1} = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-2r-r} = {}^9C_r \frac{3^{9-2r}}{2^{9-r}} (-1)^r x^{18-3r}$.To find the constant term in the product $(1+2x-3x^3)(\dots)$,we need the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ from the expansion of $(\frac{3}{2}x^2-\frac{1}{3x})^9$.$1$. For $x^0$: $18-3r = 0 \implies r=6$. Coefficient is ${}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{84 \cdot 27}{8 \cdot 729} = \frac{84}{8 \cdot 27} = \frac{21}{2 \cdot 27} = \frac{7}{18}$.$2$. For $x^{-1}$: $18-3r = -1$ (No integer solution for $r$).$3$. For $x^{-3}$: $18-3r = -3 \implies 3r=21 \implies r=7$. Coefficient is ${}^9C_7 (\frac{3}{2})^2 (-\frac{1}{3})^7 = 36 \cdot \frac{9}{4} \cdot (-\frac{1}{2187}) = 81 \cdot (-\frac{1}{2187}) = -\frac{1}{27}$.The constant term $p$ is obtained by: $(1 \cdot 	ext{coeff of } x^0) + (-3x^3 \cdot 	ext{coeff of } x^{-3}) = 1(\frac{7}{18}) - 3(-\frac{1}{27}) = \frac{7}{18} + \frac{1}{9} = \frac{7+2}{18} = \frac{9}{18} = \frac{1}{2}$.Therefore,$108p = 108 \cdot \frac{1}{2} = 54$.

If the constant term in the expansion of $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$ is $p$,then $108p$ is equal to:

Similar Questions

If the $(r + 1)^{th}$ term in the expansion of $\left( \sqrt[3]{\frac{a}{\sqrt{b}}} + \sqrt{\frac{b}{\sqrt[3]{a}}} \right)^{21}$ has the same power of $a$ and $b$,then the value of $r$ is

The number of rational terms in the binomial expansion of ${\left( 7^{\frac{1}{7}} + 11^{\frac{1}{11}} \right)^{711}}$ is:

If $p$ and $q$ are real numbers such that the $7^{\text{th}}$ term in the expansion of $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ is $700$,then $49p^2 =$ (in $q^2$)

If the coefficients of the three consecutive terms in the expansion of $(1+x)^n$ are in the ratio $1:5:20$,then the coefficient of the fourth term is $............$.

If $21^{\text{st}}$ and $22^{\text{nd}}$ terms in the expansion of $(1+x)^{44}$ are equal,then $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module