Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 21732 by tawa tawa last updated on 02/Oct/17

$$\left(\mathrm{i}\right)\mathrm{Find}\mathrm{the}\mathrm{first}\mathrm{three}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}{(2-\mathrm{x})}^6\mathrm{in}\mathrm{ascending}\mathrm{power}$$$$\mathrm{of}\mathrm{x}.$$$$\left(\mathrm{ii}\right)\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{for}\mathrm{which}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{term}\mathrm{in}{\mathrm{x}}^2\mathrm{in}\mathrm{the}\mathrm{expansion}$$$$(1+\mathrm{kx}){(2-\mathrm{x})}^6$$

Commented by Tikufly last updated on 02/Oct/17

$$\left(\mathrm{i}\right){(2-\mathrm{x})}^6{=}^6{\mathrm{C}}_0{2}^6{-}^6{\mathrm{C}}_1{2}^5\mathrm{x}{+}^6{\mathrm{C}}_2{2}^4{\mathrm{x}}^2{-}^6{\mathrm{C}}_3{2}^3{\mathrm{x}}^3+....$$$$=64-192\mathrm{x}+{240\mathrm{x}}^2-{160\mathrm{x}}^3.....$$$$\mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{terms}\mathrm{are}:-$$$$-192\mathrm{x}+{240\mathrm{x}}^3-{160\mathrm{x}}^3$$$$$$$$\left(\mathrm{ii}\right)(1+\mathrm{kx}){(2-\mathrm{x})}^6$$$$=(1+\mathrm{kx})(64-192\mathrm{x}+{240\mathrm{x}}^2-{160\mathrm{x}}^3.......)$$$$$$$$\mathrm{Since}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{term}\mathrm{of}{\mathrm{x}}^2$$$$\mathrm{Coefficient}\mathrm{of}{\mathrm{x}}^2=0$$$$=>240-192\mathrm{k}=0$$$$=>\mathrm{k}=\frac{240}{192}$$$$=>\mathrm{k}=\frac{5}{4}$$

Commented by tawa tawa last updated on 02/Oct/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com