<s>[INST] $S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + ...$ In one of my lecturer's problem sheets we were asked to evaluate the following sums:
$$S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + \dots $$
$$S2 = {x^1 \over 1!} +{x^4 \over 4!} +{x^7 \over 7!} + \dots$$
$$S3 = {x^2 \over 2!} +{x^5 \over 5!} +{x^8 \over 8!} + \dots$$
In case it's relevant, we were previously required to solve $z^3 -1 =0$, which is simple. And show that if ω is one of the complex roots of the above equation, $ω^2 + ω +1 = 0$.
How would one go about solving these sums? Is there an obvious method which I am missing?
 [/INST] Hint: We have 
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$ and 
$$e^{\omega x}=1+\omega x+\frac{1}{2!}\omega^2 x^2+\frac{1}{3!}x^3+\cdots$$ and $$e^{\omega^2 x}=1+\omega^2 x+\frac{1}{2!}\omega x^2+\frac{1}{3!}x^3+\cdots.$$
Add.  After finding $S_1(x)$ we can get the others by differentiating.
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