In the expansion of frac{e^{7x} + e^{3x}}{e^{ | vedclass

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(C) Given expression: $\frac{e^{7x} + e^{3x}}{e^{5x}} = \frac{e^{7x}}{e^{5x}} + \frac{e^{3x}}{e^{5x}} = e^{2x} + e^{-2x}$.We know the expansion of $e^

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$2$

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(C) Given expression: $\frac{e^{7x} + e^{3x}}{e^{5x}} = \frac{e^{7x}}{e^{5x}} + \frac{e^{3x}}{e^{5x}} = e^{2x} + e^{-2x}$.We know the expansion of $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$.For $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots \quad (i)$For $e^{-2x} = 1 - 2x + \frac{(-2x)^2}{2!} + \frac{(-2x)^3}{3!} + \dots = 1 - 2x + \frac{(2x)^2}{2!} - \frac{(2x)^3}{3!} + \dots \quad (ii)$Adding $(i)$ and $(ii)$:$e^{2x} + e^{-2x} = (1 + 1) + (2x - 2x) + \left(\frac{(2x)^2}{2!} + \frac{(2x)^2}{2!}\right) + \dots$$e^{2x} + e^{-2x} = 2 + 0 + 2 \cdot \frac{4x^2}{2} + \dots = 2 + 4x^2 + \dots$The constant term is the term independent of $x$,which is $2$.

In the expansion of $\frac{e^{7x} + e^{3x}}{e^{5x}}$,the constant term is

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