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MISCELLANEOUS EXERCISE 4 (II) · Q127

Q.(a+bx)(1−x)6=3−20x+cx2+…(a + bx) (1 - x)^6 = 3 -20x + cx^2 + \ldots, then find aa, bb, cc.

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(1−x)6=1−6x+15x2−⋯(1-x)^6=1-6x+15x^2-\cdots. (a+bx)(1−6x+15x2−⋯ )=a+(−6a+b)x+(15a−6b)x2+⋯(a+bx)(1-6x+15x^2-\cdots)=a+(-6a+b)x+(15a-6b)x^2+\cdots. Matching to 3−20x+cx2+⋯3-20x+cx^2+\cdots: constant: a=3a=3. xx-coefficient: −6(3)+b=−20⇒b=−2-6(3)+b=-20\Rightarrow b=-2. …

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