The expansion of e7x-exe4x is equal to

E^7x e^xe^4x =e^3x e^ 3x =[1+3x1!+(3x)^22!+...] [1 3x1!+(3x)^22! ...] =2[3x1!+(3x)^33!+...] ...

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Question

The expansion of $\frac{e^{7 x} - e^{x}}{e^{4 x}}$ is equal to

2 [1 + \frac{(3 x)^{2}}{2!} + \frac{(3 x)^{4}}{4!} + . . .]

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2 [\frac{3 x}{1!} + \frac{(3 x)^{3}}{3!} + . . .]

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2 [1 + \frac{(2 x)^{2}}{2!} + \frac{(2 x)^{4}}{4!} + . . .]

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2 [\frac{2 x}{1!} + \frac{(2 x)^{3}}{3!} + . . .]

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Solution

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Similar questions

x^{n}

\frac{e^{7 x} + e^{x}}{e^{3 x}}

(e^{x} + e^{- x})^{2} - (e^{x} - e^{- x})^{2}

e^{4 x} - e^{3 x} - 4 e^{2 x} - e^{x} + 1 = 0

\int \frac{e^{3 x} + e^{x}}{e^{4 x} - e^{2 x} + 1} d x =

\frac{e^{4 x} - 1}{2 e^{2 x}}

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