I would suggest not using summation notation for a while. Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots.$ Then $f''(x)=(2)(1)a_2+(3)(2)a_3 x+(4)(3)a_4x^2+(5)(4)a_5x^3+\cdots.$ The power series expansion of $f''(x)+f(x)$ is easy to write down using the above equations. Note that all the coefficients must be $0$.
We can't say anything about $a_0$ or $a_1$. But the constant term in the expansion of $f''(x)+f(x)$ must be $0$. Thus $a_0+(2)(1)a_2=0,$ so $a_2=-\dfrac{a_0}{(2)(1)}$.
The coefficient of $x$ in the expansion of $f''(x)+f(x)$ must be $0$. Thus $a_1+(3)(2)a_3=0,$ and therefore $a_3=-\dfrac{a_1}{(3)(2)}$.
The coefficient of $x^2$ must be $0$. So $a_2+(4)(3)a_4=0.$ Thus $a_4=-\dfrac{a_2}{(4)(3)}=\dfrac{a_0}{(4)(3)(2)(1)}$.
The coefficient of $x^3$ must be $0$. So $a_3+(5)(4)a_5=0.$ Thus $a_5=-\dfrac{a_3}{(5)(4)}=\dfrac{a_1}{(5)(4)(3)(2)}$.
I will stop here. You may want to calculate $a_6$. What is going on will become clear. Then you may want to start using general indices.