If $f\left(x\right)=$ $\sum_{n=0}^{\infty}$ $\frac{x^{3n}}{\left(3n\right)!}$ then prove $f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$

Question

QuestionIf $f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$ then prove that

$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$

My Approach$f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$

$f'\left(x\right)=$$\sum_{n=1}^{\infty}$$\frac{x^{3n-1}}{\left(3n-1\right)!}$ =$\sum_{n=0}^{\infty}$$\frac{x^{3n+2}}{\left(3n+2\right)!}$

$f''\left(x\right)=$$\sum_{n=2}^{\infty}$$\frac{x^{3n-2}}{\left(3n-2\right)!}$ =$\sum_{n=0}^{\infty}$$\frac{x^{3n+4}}{\left(3n+4\right)!}$

$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(3n\right)!}+\sum_{n=2}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!}+\sum_{n=0}^{\infty}\frac{x^{3n+4}}{\left(3n+4\right)!}$

Now i am unable to prove it to be equal to $e$$^{x}$

Answer 1

Note that we can write: $$f(x) = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \ldots $$ and $$f'(x) = \sum_{n=0}^{\infty} \frac{x^{3n+2}}{(3n+2)!}$$ $$ = \frac{x^2}{2!} + \frac{x^5}{5!} + \ldots$$ $$f''(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)!} $$ $$= x + \frac{x^4}{4!} +\ldots$$

Add them up to get $e^x$.

Answer 2

Your solution is generally ok, but for $f''$ we should rewrite it like below:
$f''=(f')'=(\sum_{n=0}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!})'=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{\left(3n+1\right)!}$

So finally, we will have:
$f\left(x\right)+f'\left(x\right)+f''\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(3n\right)!}+\frac{x^{3n+2}}{\left(3n+2\right)!}+\frac{x^{3n+1}}{\left(3n+1\right)!}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=e^x$

Note that as we had all $3n, 3n+1$ and $ 3n+2$, we could merge those 3 sigmas.