Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$
Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$
It is known $(r+1)^{\text {th }}$ term, $T_{r+1}$, in the binomial expansion of $(a+b)^{n}$ is given by ${T_{r + 1}} = {\,^n}{C_r}{a^{n - r}}{b^r}$
Thus, the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}$ is
${T_{13}} = {T_{11 + 1}} = {\,^{18}}{C_{12}}{(9x)^{18 - 12}}{\left( { - \frac{1}{{3\sqrt x }}} \right)^{12}}$
$=(-1)^{12} \frac{18 !}{1216 !}(9)^{6}(x)^{6}\left(\frac{1}{3}\right)^{12}\left(\frac{1}{\sqrt{x}}\right)^{12}$
$=\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 !}{121 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \cdot x^{6}\left(\frac{1}{x^{6}}\right) \cdot 3^{12}\left(\frac{1}{3^{12}}\right)$ $\left[9^{6}=\left(3^{2}\right)^{6}=3^{12}\right]$
$=18564$
Similar Questions
The number of terms in the expansion of ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$, which are integers is
The number of terms in the expansion of ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$, which are integers is
If the second, third and fourth term in the expansion of ${(x + a)^n}$ are $240, 720$ and $1080$ respectively, then the value of $n$ is
If the second, third and fourth term in the expansion of ${(x + a)^n}$ are $240, 720$ and $1080$ respectively, then the value of $n$ is
If in the expansion of ${(1 + x)^{21}}$, the coefficients of ${x^r}$ and ${x^{r + 1}}$ be equal, then $r$ is equal to
If in the expansion of ${(1 + x)^{21}}$, the coefficients of ${x^r}$ and ${x^{r + 1}}$ be equal, then $r$ is equal to
Coefficient of $x^6$ in the binomial expansion ${\left( {\frac{{4{x^2}}}{3}\; - \;\frac{3}{{2x}}} \right)^9}$ is
Coefficient of $x^6$ in the binomial expansion ${\left( {\frac{{4{x^2}}}{3}\; - \;\frac{3}{{2x}}} \right)^9}$ is
In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$ if the ratio of ${7^{th}}$ term from the beginning to the ${7^{th}}$ term from the end is $\frac{1}{6}$, then $n = $
In ${\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^n}$ if the ratio of ${7^{th}}$ term from the beginning to the ${7^{th}}$ term from the end is $\frac{1}{6}$, then $n = $