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Question Number 32541 by rahul 19 last updated on 27/Mar/18
Coefficient of x^5 in the expansion of (x^2 −x−2)^5 is
$${Coefficient}\:{of}\:{x}^{\mathrm{5}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)^{\mathrm{5}} \:{is} \\ $$
Answered by MJS last updated on 27/Mar/18
=(x−2)^5 (x+1)^5 coefficients of (a±b)^5 1; ±5; 10; ±10; 5; ±1 (x−2)^5 = =x^5 −10x^4 +40x^3 −80x^2 +80x−32 (x+1)^2 = =x^5 +5x^4 +10x^3 +10x^2 +5x+1 now just multiplicate those who will give x^5 1−50+400−800+400−32=−81
$$=\left({x}−\mathrm{2}\right)^{\mathrm{5}} \left({x}+\mathrm{1}\right)^{\mathrm{5}} \\ $$$$\mathrm{coefficients}\:\mathrm{of}\:\left({a}\pm{b}\right)^{\mathrm{5}} \\ $$$$\mathrm{1};\:\pm\mathrm{5};\:\mathrm{10};\:\pm\mathrm{10};\:\mathrm{5};\:\pm\mathrm{1} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{5}} = \\ $$$$={x}^{\mathrm{5}} −\mathrm{10}{x}^{\mathrm{4}} +\mathrm{40}{x}^{\mathrm{3}} −\mathrm{80}{x}^{\mathrm{2}} +\mathrm{80}{x}−\mathrm{32} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$={x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{just}\:\mathrm{multiplicate}\:\mathrm{those} \\ $$$$\mathrm{who}\:\mathrm{will}\:\mathrm{give}\:{x}^{\mathrm{5}} \\ $$$$\mathrm{1}−\mathrm{50}+\mathrm{400}−\mathrm{800}+\mathrm{400}−\mathrm{32}=−\mathrm{81} \\ $$
Commented by rahul 19 last updated on 27/Mar/18
Thank u sir.
$${Thank}\:{u}\:{sir}. \\ $$
Answered by Tinkutara last updated on 28/Mar/18
(x^2 −x−2)^5 =Σ((5!)/(a!b!c!))(x^2 )^a (−x)^b (−2)^c =Σ((5!)/(a!b!c!))(−1)^(b+c) .2^c .x^(2a+b) a+b+c=5, 2a+b=5 There are only 3 cases for a,b,c: (1)a=0,b=5,c=0 (2)a=1,b=3,c=1 (3)a=2,b=1,c=2 Coefficient of x^(2a+b) is =Σ((5!)/(a!b!c!))(−1)^(b+c) .2^c So we get ((5!)/(0!5!0!))(−1)+((5!)/(3!))×2+((5!)/(2!2!))(−1)×4 =−1+40−120=−81
$$\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)^{\mathrm{5}} =\Sigma\frac{\mathrm{5}!}{{a}!{b}!{c}!}\left({x}^{\mathrm{2}} \right)^{{a}} \left(−{x}\right)^{{b}} \left(−\mathrm{2}\right)^{{c}} \\ $$$$=\Sigma\frac{\mathrm{5}!}{{a}!{b}!{c}!}\left(−\mathrm{1}\right)^{{b}+{c}} .\mathrm{2}^{{c}} .{x}^{\mathrm{2}{a}+{b}} \\ $$$${a}+{b}+{c}=\mathrm{5},\:\mathrm{2}{a}+{b}=\mathrm{5} \\ $$$${There}\:{are}\:{only}\:\mathrm{3}\:{cases}\:{for}\:{a},{b},{c}: \\ $$$$\left(\mathrm{1}\right){a}=\mathrm{0},{b}=\mathrm{5},{c}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right){a}=\mathrm{1},{b}=\mathrm{3},{c}=\mathrm{1} \\ $$$$\left(\mathrm{3}\right){a}=\mathrm{2},{b}=\mathrm{1},{c}=\mathrm{2} \\ $$$${Coefficient}\:{of}\:{x}^{\mathrm{2}{a}+{b}} \:{is} \\ $$$$=\Sigma\frac{\mathrm{5}!}{{a}!{b}!{c}!}\left(−\mathrm{1}\right)^{{b}+{c}} .\mathrm{2}^{{c}} \\ $$$${So}\:{we}\:{get}\:\frac{\mathrm{5}!}{\mathrm{0}!\mathrm{5}!\mathrm{0}!}\left(−\mathrm{1}\right)+\frac{\mathrm{5}!}{\mathrm{3}!}×\mathrm{2}+\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{2}!}\left(−\mathrm{1}\right)×\mathrm{4} \\ $$$$=−\mathrm{1}+\mathrm{40}−\mathrm{120}=−\mathrm{81} \\ $$
Commented by rahul 19 last updated on 28/Mar/18
thank u so much.
$${thank}\:{u}\:{so}\:{much}. \\ $$

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