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Question Number 85817 by jagoll last updated on 25/Mar/20

what is coefficient of x^2 in the expansion [ (1−x)(1+2x)]^6

$$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\left[\:\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{2x}\right)\right]^{\mathrm{6}} \\ $$

Answered by john santu last updated on 25/Mar/20

(1+x−2x^2 )^(6 ) = Σ_(a+b+c≤ 6) ^6 ((( 6)),((a,b,c)) ) 1^a x^b (−2x^2 )^c ⇒ a+b+2c = 6 (i) c = 1 ∧ a = 5 ⇒ ((6!)/(5!.1!)) .1^5 (−2x^2 )^1 = −12x^2 (ii) b = 2 ∧ a = 4 ⇒ ((6!)/(4!.2!)) 1^4 .x^2 = 15x^2 hence coefficient x^2 = −12+15 = 3

$$\left(\mathrm{1}+{x}−\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{6}\:\:} =\: \\ $$$$\underset{{a}+{b}+{c}\leqslant\:\mathrm{6}} {\overset{\mathrm{6}} {\sum}}\:\begin{pmatrix}{\:\:\:\:\:\mathrm{6}}\\{{a},{b},{c}}\end{pmatrix}\:\mathrm{1}^{{a}} {x}^{{b}} \:\left(−\mathrm{2}{x}^{\mathrm{2}} \right)^{{c}} \\ $$$$\Rightarrow\:{a}+{b}+\mathrm{2}{c}\:=\:\mathrm{6} \\ $$$$\left({i}\right)\:{c}\:=\:\mathrm{1}\:\wedge\:{a}\:=\:\mathrm{5}\: \\ $$$$\Rightarrow\:\frac{\mathrm{6}!}{\mathrm{5}!.\mathrm{1}!}\:.\mathrm{1}^{\mathrm{5}} \:\left(−\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{1}} \:=\:−\mathrm{12}{x}^{\mathrm{2}} \\ $$$$\left({ii}\right)\:{b}\:=\:\mathrm{2}\:\wedge\:{a}\:=\:\mathrm{4}\: \\ $$$$\Rightarrow\:\frac{\mathrm{6}!}{\mathrm{4}!.\mathrm{2}!}\:\mathrm{1}^{\mathrm{4}} .{x}^{\mathrm{2}} \:=\:\mathrm{15}{x}^{\mathrm{2}} \\ $$$${hence}\:{coefficient}\:{x}^{\mathrm{2}} \:=\:−\mathrm{12}+\mathrm{15}\:=\:\mathrm{3} \\ $$

Commented by jagoll last updated on 25/Mar/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 25/Mar/20

(1+x−2x^2 )^6 coef. of x^2 =((6×5)/2)+6×(−2)=3

$$\left(\mathrm{1}+{x}−\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{6}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{2}} =\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}+\mathrm{6}×\left(−\mathrm{2}\right)=\mathrm{3} \\ $$

Commented by jagoll last updated on 25/Mar/20

this is short cut sir?

$$\mathrm{this}\:\mathrm{is}\:\mathrm{short}\:\mathrm{cut}\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 25/Mar/20

each of the 6 x′s will be multiplicated each of the other x′s to form x^2 , there are C_2 ^6 combinations, sum of the coef. is ((6×5)/2). each of the 6 (−2x^2 ) will be multiplicated the other 1′s, the sum of the coef. is 6×(−2). therefore ⇒((6×5)/2)+6×(−2)=3.

$${each}\:{of}\:{the}\:\:\mathrm{6}\:{x}'{s}\:{will}\:{be}\:{multiplicated} \\ $$$${each}\:{of}\:{the}\:{other}\:{x}'{s}\:{to}\:{form}\:{x}^{\mathrm{2}} , \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{6}} \:\:{combinations},\:{sum}\:{of} \\ $$$${the}\:{coef}.\:{is}\:\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}. \\ $$$${each}\:{of}\:{the}\:\mathrm{6}\:\left(−\mathrm{2}{x}^{\mathrm{2}} \right)\:{will}\:{be}\:{multiplicated} \\ $$$${the}\:{other}\:\mathrm{1}'{s},\:{the}\:{sum}\:{of}\:{the}\:{coef}.\:{is} \\ $$$$\mathrm{6}×\left(−\mathrm{2}\right). \\ $$$${therefore}\:\Rightarrow\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}+\mathrm{6}×\left(−\mathrm{2}\right)=\mathrm{3}. \\ $$

Commented by jagoll last updated on 25/Mar/20

in generally sir if the question (1+x−2x^2 )^(n ) for n even ?

$$\mathrm{in}\:\mathrm{generally}\:\mathrm{sir}\: \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{question}\:\left(\mathrm{1}+\mathrm{x}−\mathrm{2x}^{\mathrm{2}} \right)^{\mathrm{n}\:} \\ $$$$\mathrm{for}\:\mathrm{n}\:\mathrm{even}\:?\: \\ $$

Commented by jagoll last updated on 25/Mar/20

C_2 ^n + (−2n) sir

$$\mathrm{C}_{\mathrm{2}} ^{\mathrm{n}} \:+\:\left(−\mathrm{2n}\right)\:\mathrm{sir} \\ $$

Commented by mr W last updated on 25/Mar/20

n is any +ve integer, mustn′t be even.

$${n}\:{is}\:{any}\:+{ve}\:{integer},\:{mustn}'{t}\:{be}\:{even}. \\ $$

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