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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-

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

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