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Question Number 124173 by bramlexs22 last updated on 01/Dec/20

Answered by bobhans last updated on 01/Dec/20

$$(1+kx){(1-2x)}^5=(1+kx)\left[\underset{n=0}{\sum ^5}{C}_n^5{(-2x)}^{5-n}\right]$$$$=(1+kx)(C_3^5{(-2x)}^2+C{}_2^5{(-2x)}^3)$$$$=(1+kx)(80x^2-160x^3)$$$$coefficientof{x}^3\Rightarrow 80k-160=20$$$$\Rightarrow 80k=180;k=\frac{18}{8}=\frac{9}{4}$$

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{please}\mathrm{post}\mathrm{a}\mathrm{new}\mathrm{question}\mathrm{instead}\mathrm{of}\mathrm{comments}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy},\mathrm{just}\mathrm{type}\mathrm{on}\mathrm{the}\equiv \mathrm{sign}\mathrm{in}\mathrm{the}\mathrm{left}$$$$\mathrm{upper}\mathrm{corner},\mathrm{then}\mathrm{select}forum\mathrm{and}\mathrm{finally}$$$$\mathrm{the}+\mathrm{on}\mathrm{top}.$$

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