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$\large\frac{1}{e^{3x}}$$(e^x+e^{5x})=a_0+a_1x+a_2x^2+.........=>\; 2a_1+2^3a_3+2^5a_5+........=$

\[\begin {array} {1 1} (1)\;e & \quad (2)\;e^{-1} \\ (3)\;1 & \quad (4)\;0 \end {array}\]

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answered Nov 7, 2013 by pady_1
 
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