calculate-n-0-1-3n-

Question Number 166660 by mnjuly1970 last updated on 24/Feb/22

c a l c u l a t e

Ω = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = ?

Answered by amin96 last updated on 24/Feb/22

\underset{1}{\underset{⏟}{e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}}} \underset{2}{\underset{⏟}{e^{xk} = \underset{n = 0}{\sum^{\infty}} \frac{{(x k)}^{n}}{n!}}} \underset{3}{\underset{⏟}{e^{{x k}^{2}} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n} k^{2 n}}{n!}}}

(1) + (2) + (3) = 3 \times \underset{S}{\underset{⏟}{(\underset{n = 0}{\sum^{\infty}} \frac{x^{3 n}}{(3 n)!})}}

S = \frac{e^{x} + e^{k x} + e^{k^{2} x}}{3} k = \frac{- 1 + i \sqrt{3}}{2} k^{2} = \frac{- 1 - i \sqrt{3}}{2}

S = \frac{1}{3} \times (e^{x} + e^{x \times \frac{- 1 + i \sqrt{3}}{2}} + e^{x \times \frac{- 1 - i \sqrt{3}}{2}}) =

= \frac{1}{3} \times (e^{x} + e^{- \frac{x}{2}} (e^{\frac{x i \sqrt{3}}{2}} + e^{- \frac{x i \sqrt{3}}{2}})) = \frac{1}{3} \times (e^{x} + 2 e^{- \frac{x}{2}} c o s (\frac{x \sqrt{3}}{2}))

x = 1

S = \underset{n = 0}{\sum^{\infty}} \frac{1}{(3 n)!} = \frac{1}{3} \times (e + 2 e^{- \frac{1}{2}} c o s (\frac{\sqrt{3}}{2})) by MATH . AMIN

Commented by mnjuly1970 last updated on 25/Feb/22

b r a v o s i r