Question

Find the term independent of x in the expansion of the following expressions:
(i) ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$

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

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

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

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

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

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

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

(ix) ${\left(\sqrt[3]{x}+\frac{1}{2\sqrt[3]{x}}\right)}^{18},x>0$

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

Answer 1

(i) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9 \\ {T}_{r+1}={}^{9}C_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r \\ =(-1{)}^r{}^{9}C_r.\frac{3^{9-2r}}{2^{9-r}}\times x^{18-2r-r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 18-3r=0 \\ \Rightarrow 3r=18 \\ \Rightarrow r=6 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}7\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {}^{9}C_6\times \frac{3^{9-12}}{2^{9-6}} \\ =\frac{9\times 8\times 7}{3\times 2}\times 3^{-3}\times 2^{-3} \\ =\frac{7}{18} \end{gathered}$$

(ii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(2x+\frac{1}{3x^2}\right)}^9 \\ {T}_{r+1}={}^{9}C_r(2x{)}^{9-r}{\left(\frac{1}{3x^2}\right)}^r \\ ={}^{9}C_r.\frac{2^{9-r}}{3^r}{x}^{9-r-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 9-3r=0 \\ \Rightarrow r=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}4\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {}^{9}C_3\frac{2^6}{3^3} \\ ={}^{9}C_3\times \frac{64}{27} \end{gathered}$$

(iii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(2x^2-\frac{3}{x^3}\right)}^{25} \\ {T}_{r+1}={}^{25}C_r(2x^2{)}^{25-r}{\left(\frac{-3}{x^3}\right)}^r \\ =(-1{)}^r{}^{25}C_r\times 2^{25-r}\times 3^r{x}^{50-2r-3r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 50-5r=0 \\ \Rightarrow r=10 \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}11\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ (-1{)}^{10}{}^{25}C_{10}\times 2^{25-10}\times 3^{10} \\ ={}^{25}C_{10}(2^{15}\times 3^{10}) \end{gathered}$$

(iv) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(3x-\frac{2}{x^2}\right)}^{15} \\ {T}_{r+1}={}^{15}C_r(3x{)}^{15-r}{\left(\frac{-2}{x^2}\right)}^r \\ =(-1{)}^r{}^{15}C_r\times 3^{15-r}\times 2^r{x}^{15-r-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 15-3r=0 \\ \Rightarrow r=5 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}6\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ (-1{)}^5{}^{15}C_5.3^{15-5}.2^5 \\ =-3003\times 3^{10}\times 2^5 \end{gathered}$$

(v) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(\sqrt{\frac{x}{3}}+\frac{3}{2x^2}\right)}^{10} \\ {T}_{r+1}={}^{10}C_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}{\left(\frac{3}{2x^2}\right)}^r \\ ={}^{10}C_r.\frac{3^{r-{\displaystyle \frac{10-r}{2}}}}{2^r}{x}^{\frac{10-r}{2}-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{10-r}{2}-2r=0 \\ \Rightarrow 10-5r=0 \\ \Rightarrow r=2 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}3\mathrm{rd}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {}^{10}C_2\times \frac{3^{2-{\displaystyle \frac{10-2}{2}}}}{2^2} \\ =\frac{10\times 9}{2\times 4\times 9} \\ =\frac{5}{4} \end{gathered}$$

(vi) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(x-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.\right)}^{3n} \\ {T}_{r+1}={}^{3n}C_r{x}^{3n-r}{\left(\frac{-1}{x^2}\right)}^r \\ =(-1{)}^r{}^{3n}C_r{x}^{3n-r-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 3n-3r=0 \\ \Rightarrow r=n \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}(n+1)\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ (-1{)}^n{}^{3n}C_n \end{gathered}$$


(vii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(\frac{1}{2}{x}^{1/3}+x^{-1/5}\right)}^8 \\ {T}_{r+1}={}^{8}C_r{\left(\frac{1}{2}{x}^{1/3}\right)}^{8-r}(x^{-1/5}{)}^r \\ ={}^{8}C_r.\frac{1}{2^{8-r}}{x}^{\frac{8-r}{3}-\frac{r}{5}} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}\mathrm{x},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{8-r}{3}-\frac{r}{5}=0 \\ \Rightarrow 40-5r-3r=0 \\ \Rightarrow 8r=40 \\ \Rightarrow r=5 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}6\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ {}^{8}C_5\times \frac{1}{2^{8-5}} \\ =\frac{8\times 7\times 6}{3\times 2\times 8}=7 \end{gathered}$$

(ix) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(\sqrt[3]{x}+\frac{1}{2\sqrt[3]{x}}\right)}^{18} \\ {T}_{r+1}={}^{18}C_r(x^{1/3}{)}^{18-r}{\left(\frac{1}{2x^{1/3}}\right)}^r \\ ={}^{18}C_r\times \frac{1}{2^r}{x}^{\frac{18-r}{3}-\frac{r}{3}} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}r,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{18-r}{3}-\frac{r}{3}=0 \\ \Rightarrow 18-2r=0 \\ \Rightarrow r=9 \\ \mathrm{The}\mathrm{term}\mathrm{is} \\ {}^{18}C_9\times \frac{1}{2^9} \end{gathered}$$

(x) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
$$\begin{gathered} {\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^6 \\ {T}_{r+1}={}^{6}C_r{\left(\frac{3}{2}{x}^2\right)}^{6-r}{\left(\frac{-1}{3x}\right)}^r \\ ={\left(-1\right)}^r{}^{6}C_r\times \frac{3^{6-r-r}}{2^{6-r}}{x}^{12-2r-r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}x,\mathrm{we}\mathrm{must}\mathrm{have} \\ 12-3r=0 \\ \Rightarrow r=4 \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{term}\mathrm{is}\mathrm{the}4\mathrm{th}\mathrm{term}. \\ {}^{6}C_4\times \frac{3^{6-4-4}}{2^{6-4}} \\ =\frac{6\times 5}{2\times 1\times 4\times 9} \\ =\frac{5}{12} \end{gathered}$$