i) From MF26, we know that the Maclaurin expansion of the following are :
e^x = 1 + x + x²/2! + x³/3! + ...
ln (1 + x) = x - x²/2 + x³/3 - ...
We only want up to x³.
e^(2x) ln (1 + ax)
= (1 + 2x + (2x)²/2! + ...)(ax - (ax)²/2 + (ax)³/3 - ...)
= (1 + 2x + 2x² + ...)(ax - a²x²/2 + a³x³/3 - ...)
= ax - a²x²/2 + a³x³/3 + 2ax² - a²x³ + 2ax³ + ...
= ax + (2a - a²/2)x² + (a³/3 - a² + 2a)x³ + ...
= ax + a(2 - a/2)x² + a(a²/3 - a + 2)x³ + ...

For there to be no term in x², the coefficient of the x² term must be 0.
So, a(2 - a/2) = 0
a = 0 (N.A since a is non-zero) or 2 - a/2 = 0
2 = a/2
a = 4


part ii) in the next post.