sum o series....

\\\texttt{let S be our required sum } \\ \\\texttt{consider this expansion } \\ Re(e^{\omega x})=S(x)-\frac{1}{2}\left(e^{x} \right -S(x)) \\ 2e^{\frac{-1}{2}}\cos\left( \frac{\sqrt{3}}{2}\right)=3S-e \\ \boxed{S=\frac{1}{3}\left(e+\frac{2cos\frac{\sqrt{3}}{2}}{\sqrt{e}} \right)}

cipher: try the expansion of (e+e^ω + e^ω2)/3

ull see that that's what the Q asks

Now e^ω = e^{-1/2 + i√3/2}
e^w2 = e^{-1/2 - i√3/2}

add both. it will be e^-1/2(2cos(√3/2))

So the final expression boils down to
e + e^-1/2(2cos(√32))3

which is what akari got.

(i cudnt understand his method though)

wat i meant was
if u see the expansion of :
e^{\omega x}=\left( 1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+...\infty\right)+\omega \left(x+\frac{x^4}{4!}+\frac{x^7}{7!}+....\infty \right)+\omega ^2\left(\frac{x^2}{2!}+\frac{x^5}{5!}+.........\infty \right) \\ \\ ( \because \omega ^{3n}=1,\omega ^{3n+1}=\omega,\omega ^{3n+2}=\omega ^{2}) \\ \\\texttt{now put }\omega =\frac{-1}{2}+\frac{i\sqrt{3}}{2} \ and \ \omega ^2=\frac{-1}{2}-\frac{\sqrt{3}}{2} \\ \\ \\\texttt{we will then see the real part of r.h.s is}\\ \\ Re(e^{\omega x})=\left( 1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+...\infty\right)-\frac{1}{2} \left(x+\frac{x^4}{4!}+\frac{x^7}{7!}+....\infty \right)-\frac{1}{2}\left(\frac{x^2}{2!}+\frac{x^5}{5!}+.........\infty \right) \\ \\\texttt{Put x=1 and let S be our summation} \\ Re(e^{\omega x})=S-\frac{1}{2}\left(e-S \right)