- Home
- Standard 11
- Mathematics
Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$
$18564$
$18564$
$18564$
$18564$
Solution
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
${T_{13}} = {T_{11 + 1}} = {\,^{18}}{C_{12}}{(9x)^{18 – 12}}{\left( { – \frac{1}{{3\sqrt x }}} \right)^{12}}$
$=(-1)^{12} \frac{18 !}{1216 !}(9)^{6}(x)^{6}\left(\frac{1}{3}\right)^{12}\left(\frac{1}{\sqrt{x}}\right)^{12}$
$=\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 !}{121 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \cdot x^{6}\left(\frac{1}{x^{6}}\right) \cdot 3^{12}\left(\frac{1}{3^{12}}\right)$ $\left[9^{6}=\left(3^{2}\right)^{6}=3^{12}\right]$
$=18564$
