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Exercise 5.4 · Q9

Q.Find the coefficient of x4x^4 in the expansion of 3−4x+x2e2x\dfrac{3-4x+x^2}{e^{2x}}.

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Rewrite the expression as (3−4x+x2)⋅e−2x(3-4x+x^2)\cdot e^{-2x}, and combine the relevant coefficients from the exponential series with the coefficients 3,−4,13,-4,1 of the polynomial factor.

Step 1. Rewrite. 3−4x+x2e2x=(3−4x+x2) e−2x\dfrac{3-4x+x^2}{e^{2x}} = (3-4x+x^2)\,e^{-2x}.

Step 2. Coefficients of e−2x=∑(−2x)kk!e^{-2x}=\sum\dfrac{(-2x)^k}{k!} needed.

Coefficient of x2x^2: (−2)22!=42=2\dfrac{(-2)^2}{2!}=\dfrac42=2. Coefficient of x3x^3: (−2)33!=−86=−43\dfrac{(-2)^3}{3!}=\dfrac{-8}6=-\dfrac43. Coefficient of x4x^4: (−2)44!=1624=23\dfrac{(-2)^4}{4!}=\dfrac{16}{24}=\dfrac23.

Step 3. Assemble the coefficient of x4x^4 in the product. A term x4x^4 in (3−4x+x2)⋅e−2x(3-4x+x^2)\cdot e^{-2x} arises three ways: …

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