Question

Find the term independent of x in the expasion of the following expressions:

$\left(i\right){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9\left(ii\right){(2x+\frac{1}{3x^2})}^9\left(iii\right){(2x^2-\frac{3}{3x^3})}^{25}\left(iv\right){(3x-\frac{2}{2x^2})}^{15}\left(v\right)(\sqrt{\frac{x}{3}}+{\frac{3}{2x^2}}^{10})\left(vi\right){(x-\frac{1}{x^2})}^{3n}\left(vii\right){(\frac{1}{2}{x}^{1/3}+x^{-1/5})}^8\left(viii\right)(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9\left(ix\right){(\sqrt[3]{3x}+\frac{1}{2\sqrt[3]{3x}})}^{18},x>2\left(x\right){(\frac{3}{2}{x}^2-\frac{1}{3x})}^6$

Answer 1

$\left(i\right)(\frac{3}{2}{x}^2-\frac{1}{3x})$

$T_{r+1}{=}^9{C}_r{\left(\frac{3x^2}{2}\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r$

${=}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}\left(x^{18-2r}\right){\left(\frac{-1}{3}\right)}^r{x}^{-r}$

$\text{Let T{r+1}~be~independent~of~x}$ $18-3r=0orr=6$ $\therefore \text{Required term}$

$\Rightarrow T_{r+1}=T_{6+1}=T_7{=}^9{C}_6{\left(\frac{3}{2}\right)}^{9-6}$

$=84\left(\frac{27}{8}\right)\left(\frac{1}{179}\right)x^0=\frac{7}{18}$

$\left(ii\right)(2x+\frac{1}{3x^2})$

$\text{Suppose the (r+1th)term in the given expression is independent of x}$

Now ${(2x+\frac{1}{3x^2})}^{9}4$

$T_{r+1}{=}^9{C}_r(2x{)}^{9-r}{\left(\frac{1}{3x^2}\right)}^r$

$=9C_r.\frac{2^{9-r}}{3^r}{x}^{9-r-2r}$

$\text{For this term to be independent of x, we must have 9 -3r = 0}$

$\Rightarrow r=3$

$\text{Hence, the required term is the 4th term.}$

$\text{Now, we have}$

${}^9{C}_3\frac{2^6}{3^3}{=}^9{C}_3\times \frac{64}{27}$

($iii){(2x^2-\frac{3}{x^3})}^{25}$

$T_{r+1}=(-1{)}^r{}^n{C}_r(2x^2{)}^{25-r}{\left(\frac{3}{x^3}\right)}^r$

$=(-1{)}^r{}^n{C}_{r}2x^{25-r}{3}^r{x}^{50-2r-3r}$

$\text{Term independent of}x=x^0$

$\Rightarrow x^{50-50r}=x^0$

$\Rightarrow 50-5r=0\Rightarrow r=10$

$\therefore t_{11}=(-1{)}^{1}0{}^{25C_{10}}{2}^{15}\times 3^{10}$

$={}^{25}{C}_{10}{2}^{15}{3}^{10}$

$\left(iv\right){(3x-\frac{2}{x^2})}^{15}$

$T_{r+1}=(-1{)}^r{}^{15}{C}_r(3x{)}^{15-r}{\left(\frac{2}{x^2}\right)}^r$

$=(-1{)}^r{}^{15}{C}_r{3}^{15-r}2r{x}^{15-r-1r}$

$\text{Term independent of x}\Rightarrow x^0$

$\Rightarrow x^{15-3r}=x^0$

$15-3r=0\Rightarrow r=5$

$\therefore t_6=(-1{)}^5{}^{15}{C}_5{3}^{10}{2}^5$

$=-\frac{15!}{5!10!}\times 3^{10}{2}^5$

$=-\frac{15\times 14\times 13\times 12\times 11}{120}{3}^{10}{2}^5$

$=-3003\times 3^{10}\times 2^5$

$\left(V\right){(\sqrt{\frac{x}{3}}+\frac{3}{2x^2})}^{10}$

$T_{r+1}={}^{10}{C}_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}{\left(\frac{3}{2x^2}\right)}^r$

$={}^{10}{C}_r{x}^{5-\frac{r}{2}-2r}\times 3^r\times 3^{-5+\frac{r}{2}}\times 2^{-r}$

$\text{Independent of x}\Rightarrow x^0$

$x\frac{10-r-4r}{r}=x^0$

$10-5r=0$

$r=2$

$t_3-{}^{10}{C}_2{3}^{2-5+1}{2}^{-2}$

$={}^{10}{C}_2{3}^{-2}{2}^{-2}$

$=\frac{10!}{2!8!}\times \frac{1}{36}=\frac{10\times 9}{2\times 36}=\frac{5}{4}$

$\left(vi\right){(x-\frac{1}{x^2})}^{3n}$

$\text{Suppose the}(r+1{)}^{th}\text{term in the given expression is independent of x}$

Now, ${(x-\frac{1}{x^2})}^{3n}$

$T_{r+1}={}^{3n}{C}_r{x}^{3n-r}{\left(\frac{-1}{x^2}\right)}^r$

$=(-1{)}^r{}^{3n}{C}_r{x}^{3n-r-2r}$

$\text{For this term to be independent of x, we must have 3n - 3r = 0}$

$\Rightarrow r=n$ $\text{Hence, the required term is the (n+1)th term}.$

$\text{Now, we have}$

$(-1)n{}^{3n}{C}_n$

$\left(vii\right){(\frac{1}{2}{x}^{1/3}+x^{-1/5})}^8$

$\text{We have,}$

${(\frac{1}{2}{x}^{1/3}+x^{-1/5})}^8$

$\text{Let}(r+1{)}^{th}\text{term be independent of x}$

$\therefore T_{r+1}={}^8{C}_r{\left(\frac{1}{2}{x}^{\frac{1}{3}}\right)}^{8-r}{\left(x\frac{-1}{5}\right)}^r$

$=8C_r{\left(\frac{1}{2}\right)}^{8-r}\times {\left(x^{\frac{1}{3}}\right)}^{8-r}\times {\left(\frac{1}{x^{\frac{1}{5}}}\right)}^r$

${=}^8{C}_r{\left(\frac{1}{2}\right)}^{8-r}\times (x{)}^{\frac{8-r}{3}}\times \left(\frac{1}{x^{\frac{1}{5}}}\right)$

${=}^8{C}_r{\left(\frac{1}{2}\right)}^{8-r}\times (x{)}^{\frac{8-rr}{35}}$

${=}^8{C}_r{\left(\frac{1}{2}\right)}^{8-r}\times (x{)}^{\frac{40-5r-3r}{15}}$

${=}^8{C}_r{\left(\frac{1}{2}\right)}^{8-r}\times (x{)}^{\frac{40-8r}{15}}$

$\text{It it is independent of x, we must have}$ $=\frac{40-8r}{15}=0$

$\Rightarrow 8r=40$

$\Rightarrow r=5$

$\therefore \text{The term independent of}x=T_6$

$Now$

$T_6{=}^8{C}_r{\left(\frac{1}{2}{x}^{\frac{1}{3}}\right)}^{8-5}{\left(x^{\frac{-1}{5}}\right)}^5$

$=56\times {\left(\frac{1}{2}\right)}^3$

$=56\times \frac{1}{8}$

$=7$ $\text{Hence, required term = 7}.$

$\left(viii\right)(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$

$=(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$

$=(1+x+2x^3)$

$\left[\begin{array}{cc}{\left(\frac{3}{2}{x}^2\right)}^9{-}^9{C}_1& {\left(\frac{3}{2}{x}^2\right)}^8\frac{1}{3x}...{+}^9{C}_6{\left(\frac{3}{2}{x}^2\right)}^3\\ & \left({\frac{1}{3x}}^6{-}^9{C}_7{\left(\frac{3}{2}{x}^2\right)}^2\left({\frac{1}{3x}}^7\right)\right)\end{array}\right]$

In the second bracket, we have to search the term so $x^0$ and $\frac{1}{x^3}$ which when multiplying by 1 and $2x^3$ is firs bracket will give the term in dependent of x. The term containing $\frac{1}{x}$ will not occur is second bracket. The term independent of x $=[{}^9{C}_6\frac{3^3}{2^3}\times \frac{1}{3^6}]-2x^3[{}^9{C}_7\frac{3^3}{2^3\times \frac{1}{3^7}}\times \frac{1}{x^3}]$

$=[\frac{9\times 8\times 7}{1\times 2\times 3}\times \frac{1}{8\times 27}]-2[\frac{9\times 8}{1\times 2}-\frac{1}{4\times 243}]$ $=\frac{7}{18}-\frac{2}{27}$

$=\frac{17}{54}$

$\text{Required term}=\frac{17}{54}.$

$\left(ix\right){(\sqrt[3]{3x}+\frac{1}{2\sqrt[3]{3x}})}^{18},x>2\text{We have},$

${(\sqrt[3]{3x}+\frac{1}{\sqrt[3]{3x}})}^{18},x>0$

$Let(r+1)th\text{term be independent of x}.$

$\therefore T_{r+1}={}^{18}{C}_r{\left(\sqrt[3]{3x}\right)}^{18-r}\times {\left(\frac{1}{2\sqrt[3]{3x}}\right)}^r$

$={}^{18}{C}_r{\left(\right(x{)}^{\frac{1}{3}})}^{18-r}\times {\left(\frac{1}{2}\right)}^r\times {\left(\frac{1}{x^{\frac{1}{3}}}\right)}^r$ $={}^{18}{C}_r(x{)}^{\frac{18-r}{3}}\times \left(\frac{1}{x^{\frac{r}{3}}}\right)\times {\left(\frac{1}{2}\right)}^r$

$={}^{18}{C}_r(x{)}^{\frac{18-r}{3}}\times {\left(\frac{1}{2}\right)}^r$

$={}^{18}{C}_r(x{)}^{\frac{18-r}{3}}\times {\left(\frac{1}{2}\right)}^r$

If it is independent of x, we must have $\frac{18-2r}{3}=0$

$\Rightarrow 18=2r$ $\Rightarrow r=9$

$\therefore \text{Term independent of}x=T_{9+1}=T_{10}$

Now, $T_{10}={}^{18}{C}_9{\left(\sqrt[3]{3x}\right)}^{18-9}{\left(\frac{1}{2\sqrt[3]{3x}}\right)}^9$

$={}^{18}{C}_9{\left(\sqrt[3]{3x}\right)}^9\times \frac{1}{2^9}\times {\left(\frac{1}{\sqrt[3]{3x}}\right)}^9$

$=\frac{18C_9}{2^9}$

$\text{Hence, required term =}\frac{18C_9}{2^9}.$

$\left(x\right){(\frac{3}{x}{x}^2-\frac{1}{3x})}^6$

$\text{In expansion}$

$T_{r+1}{=}^6{C}_r{\left(\frac{3x^2}{2}\right)}^{6-2}{(-\frac{1}{3x})}^r$

${=}^6{C}_r{\left(\frac{3}{2}\right)}^{6-2}\left(x^{12-3r}\right){(-\frac{1}{3})}^r$

$Let{T}_{r+1}\text{be independent of x,}$

$12-3r-0orr=4$

$\therefore \text{Required term}$

$\Rightarrow T_{r+1}=T_{4+1}=T_5{=}^6{C}_4{\left(\frac{3}{2}\right)}_{x^{12-3\left(4\right)}}^{6-4}$

$=15\left(\frac{9}{4}\right)\left(\frac{1}{18}\right)x^{\circ }=\frac{5}{12}$