Finding the second Taylor polynomial for $f(x)$ given as a series

Question

I came across this question when preparing for my midterm and I can't figure out how to solve it. The first part of the question asks to combine 2 series to determine $f(x)$, which I found to be:
$$f(x) = \sum_{k=1}^n (\frac{2}{k}-3.4^k)x^k$$
The question then asks to find the second Taylor polynomial for $f(x)$ based at $b = 0$. I'm familiar with finding the Taylor polynomial for a normal function, but I've never seen anything like this before. Any help would be appreciated!
Thanks for your time.

Answer 1

The second Taylor polynomial for $f(x)$ at $x_0$ is given by $$ f(x) \approx f(x_0) +f'(x_0) (x-x_0) + \dfrac{f''(x_0)}{2!}(x-x_0)^2. $$ Now, think about your first and second derivative of $f$: $$ f'(x)=\left(2-3.4^1 \right) + \left(2 -3.4^2.2 \right)x + \left(2-3.4^3.3 \right)x^2 + \cdots + \left(2-3.4^n.n \right)x^{n-1} $$ and $$ f''(x) = \left(2.1 -3.4^2.2.1 \right) + \left(2.2-3.4^3.3.2 \right)x + \cdots + \left(2.(n-1)-3.4^n.n.(n-1) \right)x^{n-2}, $$ i.e. $$ f'(x) = \sum_{k=1}^{n} \left( 2 - 3.4^k.k\right)x^{k-1}\;\;\mbox{and}\;\; f''(x) = \sum_{k=2}^{n} \left( 2.(k-1) - 3.4^k.k.(k-1)\right)x^{k-2}. $$