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

(i) Find the first three terms in the expansion of (2 − x)^6 in ascending power of x. (ii) Find the value of k for which there is no term in x^2 in the expansion (1 + kx)(2 − x)^6

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

Commented by Tikufly last updated on 02/Oct/17

(i) (2−x)^6 =^6 C_0 2^6 −^6 C_1 2^5 x+^6 C_2 2^4 x^2 −^6 C_3 2^3 x^3 +.... =64−192x+240x^2 −160x^3 ..... Hence, the required terms are:− −192x+240x^3 −160x^3 (ii) (1+kx)(2−x)^6 =(1+kx)(64−192x+240x^2 −160x^3 .......) Since there is no term of x^2 Coefficient of x^2 =0 => 240−192k=0 => k=((240)/(192)) => k=(5/4)

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

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

God bless you sir.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com