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Question Number 95888 by i jagooll last updated on 28/May/20

$${(1-2\mathrm{x})}^5{(2+\mathrm{x})}^6=\mathrm{a}+\mathrm{bx}+{\mathrm{cx}}^2+{\mathrm{dx}}^3+...$$$$\mathrm{find}:\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{and}\mathrm{d}$$

Answered by john santu last updated on 28/May/20

$$\Rightarrow (2+\mathrm{x})\underset{\mathrm{k}=0}{\sum ^5}\mathrm{C}(5,\mathrm{k}){\left(2\right)}^{5-\mathrm{k}}{(-(3\mathrm{x}+{2\mathrm{x}}^2))}^{\mathrm{k}}$$$$\mathrm{clearly}\mathrm{a}=64$$$$\mathrm{term}\mathrm{containing}\mathrm{x}\Rightarrow (2+\mathrm{x})\mathrm{C}(5,1)2^4(-3\mathrm{x}-{2\mathrm{x}}^2)$$$$=80(2+\mathrm{x})(-3\mathrm{x}-{2\mathrm{x}}^2)$$$$=-480\mathrm{x}-{560\mathrm{x}}^2-{160\mathrm{x}}^3$$$$\mathrm{so}\mathrm{b}=-480$$$$\mathrm{term}\mathrm{containing}{\mathrm{x}}^2$$$$=(2+\mathrm{x})\mathrm{C}(5,2)2^3{(3\mathrm{x}+{2\mathrm{x}}^2)}^2$$$$=80(2+\mathrm{x})({9\mathrm{x}}^2+{12\mathrm{x}}^3+{4\mathrm{x}}^4)$$$$\mathrm{so}\mathrm{term}{\mathrm{x}}^2={560\mathrm{x}}^2+{1440\mathrm{x}}^2$$$$\mathrm{then}\mathrm{we}\mathrm{get}\mathrm{c}=2000$$$$\mathrm{term}\mathrm{containing}{\mathrm{x}}^3$$$$=(2+\mathrm{x})\mathrm{C}(5,3)2^2{(-3\mathrm{x}-{2\mathrm{x}}^2)}^3$$$$=40(2+\mathrm{x})(-{27\mathrm{x}}^3+...)$$$$=-{2160\mathrm{x}}^3+...$$$$\mathrm{so}\mathrm{term}{\mathrm{x}}^3=-{160\mathrm{x}}^3+{1920\mathrm{x}}^3+{720\mathrm{x}}^3-{2160\mathrm{x}}^3$$$$={320\mathrm{x}}^3$$$$\mathrm{then}\mathrm{d}=320$$$$$$

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