series expansion of complex expression and its $n$ th derivatives

Question

Finding maclaurin series expansion of $$f(t)=e^{3t}+\frac{2}{1-10t^2}$$ and also find value of $f^{50}(t)$ is

What i try

Series expansion

$\displaystyle f(t)=f(0)+\frac{f'(0)}{1!}t+\frac{f''(0)}{2!}t^2+\frac{f'''(0)}{3!}t^3+\cdots +\frac{f^{(51)}(0)}{51!}t^{51}+\cdots$

Now $f(0)=1+2=3$ and $\displaystyle f'(t)=3e^{3t}+\frac{2\cdot 20t}{(1-10t^2)}$ then $f'(0)=3$

And $\displaystyle f''(t)=3^2 e^{3t}+40\bigg[\frac{(1-10t^2)^2+2t^2(1-10t^2)}{(1-10t^2)^{4}}\bigg]$

So $f''(0)=3^2+40$

How do i solve for $f'''(t)$ and for higher power of derivative . Help me please

Answer 1

If $g(t)=e^{3t}$ and $h(t)=\frac2{1-10t^2}$, then $f=g+h$. On the other hand, you have$$g(t)=\sum_{n=0}^\infty\frac{3^n}{n!}t^n$$and$$h(t)=\sum_{n=0}^\infty2\times10^nt^{2n}.$$So$$f^{(50)}(0)=3^{50}+50!\times2\times10^{25}.$$

Answer 2

You can use the following $\frac{2}{1-10t^2}=\frac{1}{1-\sqrt{10}t} + \frac{1}{1+\sqrt{10}t}$. The derivatives of these fractions are not so difficult.