NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

Ans : \((1-2 x)^{5}\)    
\(=^{5} \mathrm{C}_{0}(1)^{5}-^{5} \mathrm{C}_{1}(1)^{4}(2 \mathrm{x})+^{5} \mathrm{C}_{2}(1)^{3}(2 \mathrm{x})^{2}-^{5} \mathrm{C}_{3}(1)^{2}(2 \mathrm{x})^{3}+^{5} \mathrm{C}_{4}(1)^{1}(2 \mathrm{x})^{4}-^{5} \mathrm{C}_{3}(2 \mathrm{x})^{5}\)    
\(=1-5(2 x)+10\left(4 x^{2}\right)-10\left(8 x^{3}\right)+5\left(16 x^{4}\right)-\left(32 x^{2}\right)\)    
\(=1-10 x+40 x^{2}-80 x^{3}+80 x^{4}-32 x^{5}\) 

Ans : \(\left(\frac{2}{\mathrm{x}}-\frac{\mathrm{x}}{2}\right)^{5}=^{5} \mathrm{C}_{0}\left(\frac{2}{\mathrm{x}}\right)^{3}-^{5} \mathrm{C}_{1}\left(\frac{2}{\mathrm{x}}\right)^{4}\left(\frac{\mathrm{x}}{2}\right)+^{5} \mathrm{C}_{2}\left(\frac{2}{\mathrm{x}}\right)^{3}\left(\frac{\mathrm{x}}{2}\right)^{2}\)
\(-^{3} \mathrm{C}_{3}\left(\frac{2}{\mathrm{x}}\right)^{2}\left(\frac{\mathrm{x}}{2}\right)^{3}+^{3} \mathrm{C}_{4}\left(\frac{2}{\mathrm{x}}\right)\left(\frac{\mathrm{x}}{2}\right)^{4}-^{3} \mathrm{C}_{5}\left(\frac{\mathrm{x}}{2}\right)^{5}\)    
\(\begin{array}{l}{=\frac{32}{x^{5}}-5\left(\frac{16}{x^{4}}\right)\left(\frac{x}{2}\right)+10\left(\frac{8}{x^{3}}\right)\left(\frac{x^{2}}{4}\right)-10\left(\frac{4}{x^{2}}\right)+5\left(\frac{2}{x}\right)\left(\frac{x^{4}}{16}\right)-\frac{x^{5}}{32}} \\ {=\frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5 x+\frac{5}{8} x^{3}-\frac{x^{3}}{32}}\end{array}\) 

Ans : By using Binomial theorem,  the expression  \((2 x-3)^{6}\) can be expressed as     
\((2 x-3)^{6}=^{6} \mathrm{C}_{0}(2 x)^{6}-^{6} \mathrm{C}_{1}(2 x)^{5}(3)+^{6} \mathrm{C}_{1}(2 x)^{4}(3)^{2}-^{4} \mathrm{C}_{3}(2 x)^{3}(3)^{3}\)    
\(\begin{aligned} &=64 x^{6}-6\left(32 x^{5}\right)(3)+15\left(16 x^{4}\right)(9)-20\left(8 x^{3}\right)(27) \\ &+15\left(4 x^{2}\right)(81)-6(2 x)(243)+729 \\ &=64 x^{6}-576 x^{5}+2160 x^{4}-4320 x^{3}+4860 x^{2}-2916 x+729 \end{aligned}\) 

Ans : By using Binomial Theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\) can be expand as    
\(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}=^{5} C_{0}\left(\frac{x}{3}\right)^{5}+^{3} C_{1}\left(\frac{x}{3}\right)^{4}\left(\frac{1}{x}\right)+^{3} C_{2}\left(\frac{x}{3}\right)^{3}\left(\frac{1}{x}\right)^{2}\)    
\(\begin{array}{l}{=\frac{x^{5}}{243}+5\left(\frac{x^{4}}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^{3}}{27}\right)\left(\frac{1}{x^{2}}\right)+10\left(\frac{x^{2}}{9}\right)\left(\frac{1}{x^{3}}\right)+5\left(\frac{x}{3}\right)\left(\frac{1}{x^{4}}\right)+\frac{1}{x^{5}}} \\ {=\frac{x^{5}}{243}+\frac{5 x^{3}}{81}+\frac{10 x}{27}+\frac{10}{9 x}+\frac{5}{3 x^{3}}+\frac{1}{x^{3}}}\end{array}\) 

Ans : \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{6}=^{6} \mathrm{C}_{0}(\mathrm{x})^{6}+^{6} \mathrm{C}_{1}(\mathrm{x})^{\prime}\left(\frac{1}{\mathrm{x}}\right)+^{6} \mathrm{C}_{2}(\mathrm{x})^{4}\left(\frac{1}{\mathrm{x}}\right)^{2}\)    
\(\begin{array}{l}{+^{6} \mathrm{C}_{3}(\mathrm{x})^{3}\left(\frac{1}{\mathrm{x}}\right)^{3}+^{6} \mathrm{C}_{4}(\mathrm{x})^{2}\left(\frac{1}{\mathrm{x}}\right)^{4}+^{6} \mathrm{C}_{3}(\mathrm{x})\left(\frac{1}{\mathrm{x}}\right)^{5}+^{6} \mathrm{C}_{6}\left(\frac{1}{\mathrm{x}}\right)^{6}} \\ {=\mathrm{x}^{4}+6(\mathrm{x})^{3}\left(\frac{1}{\mathrm{x}}\right)+15(\mathrm{x})^{4}\left(\frac{1}{\mathrm{x}^{2}}\right)+20(\mathrm{x})^{3}\left(\frac{1}{\mathrm{x}^{3}}\right)+15(\mathrm{x})^{2}\left(\frac{1}{\mathrm{x}^{4}}\right)+6(\mathrm{x})\left(\frac{1}{\mathrm{x}^{5}}\right)+\frac{1}{\mathrm{x}^{6}}}\end{array}\)    
\(=x^{6}+6 x^{4}+15 x^{2}+20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}\)