Convergence Question related to $e^x$

Question

This is an IB question about power series and convergence. I have no idea how to even approach this question.

For 3a) I have figured out up to using the power series equation $$ f(x) = (1 + ax)(1 + bx) = (1 + ax)(1 − bx + \ldots (−1)^n \cdot b^nx^n +\ldots) $$ Now I have no idea how this can be turned to: $c_n = (−b)^{n-1}\cdot(a-b)$

Also for 3b) What is it even asking? Where did the $e^x$ come from? And what $x^2$?

Please help!!

Thank you

Answer 1

\begin{align} f(x) &= (1 + ax)(1 + bx) \\ &= (1 + ax)(1 − bx + \ldots +(−1)^n\cdot b^n\cdot x^n + \ldots)\\ &= (1)(1 − bx + \ldots+(−1)^n\cdot b^n\cdot x^n + \ldots) \\ &+ (ax)(1 − bx + \ldots+(−1)^n\cdot b^n\cdot x^n + \ldots)\\ &= 1 − bx + \ldots+(−1)^n\cdot b^n\cdot x^n + \ldots \\ &+ (ax − abx^2 + ...+(−1)^n \cdot a \cdot b^n\cdot x^{n+1} + ...) \end{align} Now notice that the terms involving $x^{n+1}$ are

1. $(-1)^{n+1} b^{n+1} x^{n+1} = (-1)(-1)^{n} b \cdot b^{n} x^{n+1}$, and
2. $(−1)^n \cdot a \cdot b^n\cdot x^{n+1}$

Summing gets us

\begin{align} S &= (-1)(-1)^{n} b \cdot b^{n} x^{n+1} + (−1)^n \cdot a \cdot b^n\cdot x^{n+1} \\ &= (-1)^n x^{n+1} b^n\left( (-1) b + \cdot a \right) \\ &= (-1)^n x^{n+1} b^n\left( a-b \right) \\ &= (-1)^n x^{n+1} b^n\left( a-b \right) \end{align} So the coefficient of $x^n$ (one term before $S$, hence involving one more $-$ sign) is $$ (-1)^{n-1} b^{n-1} (a - b) = (-b)^{n-1} (a-b). $$