Expand Σ ((4n−1)/3)+(2/3)Σ_(k=1) ^(n−1) cos(120k)


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Question Number 74819 by Raxreedoroid last updated on 01/Dec/19

$Expand Σ$ $\frac{4 n - 1}{3} + \frac{2}{3} \underset{k = 1}{\sum^{n - 1}} c o s (120 k)$
Commented by mathmax by abdo last updated on 01/Dec/19
$let S_n =((4n−1)/3) +(2/3)Σ_(k=1) ^(n−1) cos(120k) first let find W_n (α) =Σ_(k=1) ^(n−1) cos(αk) ⇒W_n (α)=Re(Σ_(k=1) ^(n−1) e^(iαk) ) and Σ_(k=1) ^(n−1) e^(iαk) =Σ_(k=0) ^(n−1) (e^(iα) )^k −1 =((1−e^(inα) )/(1−e^(iα) )) −1 =((1−cos(nα)−isin(nα))/(1−cos(α)−isin(α))) −1 =((2sin^2 (((nα)/2))−2isin(((nα)/2))cos(((nα)/2)))/(2sin^2 ((α/2))−2i sin((α/2))cos((α/2))))−1 =((−isin(((nα)/2))e^((inα)/2) )/(−isin((α/2))e^((iα)/2) ))−1 =((sin(((nα)/2)))/(sin((α/2))))e^(i(((n−1)α)/2))) −1 ⇒W_n (α)=((sin(((nα)/2))cos((((n−1)α)/2)))/(sin((α/2))))−1 Σ_(k=1) ^(n−1) cos(120k) =W_n (120) =((sin(60n)cos(60(n−1))/(sin(60)))−1 ⇒ S_n =((4n−1)/3)+(2/3){((sin(60n)cos(60n−60))/(sin(60)))−1}$
$l e t S_{n} = \frac{4 n - 1}{3} + \frac{2}{3} \sum_{k = 1}^{n - 1} c o s (120 k) f i r s t l e t f i n d$ $W_{n} (α) = \sum_{k = 1}^{n - 1} c o s (α k) \Rightarrow W_{n} (α) = R e (\sum_{k = 1}^{n - 1} e^{i α k})$ $a n d \sum_{k = 1}^{n - 1} e^{i α k} = \sum_{k = 0}^{n - 1} {(e^{i α})}^{k} - 1 = \frac{1 - e^{i n α}}{1 - e^{i α}} - 1$ $= \frac{1 - c o s (n α) - i s i n (n α)}{1 - c o s (α) - i s i n (α)} - 1$ $= \frac{2 {s i n}^{2} (\frac{n α}{2}) - 2 i s i n (\frac{n α}{2}) c o s (\frac{n α}{2})}{2 {s i n}^{2} (\frac{α}{2}) - 2 i s i n (\frac{α}{2}) c o s (\frac{α}{2})} - 1 = \frac{- i s i n (\frac{n α}{2}) e^{\frac{i n α}{2}}}{- i s i n (\frac{α}{2}) e^{\frac{i α}{2}}} - 1$ $= \frac{s i n (\frac{n α}{2})}{s i n (\frac{α}{2})} e^{i (\frac{n - 1) α}{2})} - 1 \Rightarrow W_{n} (α) = \frac{s i n (\frac{n α}{2}) c o s (\frac{(n - 1) α}{2})}{s i n (\frac{α}{2})} - 1$ $\sum_{k = 1}^{n - 1} c o s (120 k) = W_{n} (120) = \frac{s i n (60 n) c o s (60 (n - 1)}{s i n (60)} - 1 \Rightarrow$ $S_{n} = \frac{4 n - 1}{3} + \frac{2}{3} {\frac{s i n (60 n) c o s (60 n - 60)}{s i n (60)} - 1}$
Commented by mathmax by abdo last updated on 01/Dec/19
$i considered that 120 mean 120 radian if the mesure is in degre α =((120π)/(180)) =((2×6π)/(3×6π)) =((2π)/3) in this case we get S_n =((4n−1)/3) +(2/3){ ((sin(((nπ)/3))cos(((n−1)/3)π))/(sin((π/3))))−1}$
$i c o n s i d e r e d t h a t 120 m e a n 120 r a d i a n i f t h e m e s u r e i s i n d e g r e$ $α = \frac{120 π}{180} = \frac{2 \times 6 π}{3 \times 6 π} = \frac{2 π}{3} i n t h i s c a s e w e g e t$ $S_{n} = \frac{4 n - 1}{3} + \frac{2}{3} {\frac{s i n (\frac{n π}{3}) c o s (\frac{n - 1}{3} π)}{s i n (\frac{π}{3})} - 1}$
	Answered by mind is power last updated on 01/Dec/19

	$\cos (kx) = Re (e^{ikx})$ $\underset{k = 1}{\sum^{n - 1}} e^{ikx} = e^{ix} . \frac{(1 - {(e^{ix})}^{n - 1}}{1 - e^{ix}}$ $= \frac{e^{ix (\frac{n + 1}{2})}}{e^{i \frac{x}{2}}} . \frac{\sin (\frac{n - 1}{2}) x}{\sin (\frac{x}{2})}, Re (Σ e^{ikx}) = Σ \cos (kx) = \cos (\frac{nx}{2}) . \frac{\sin (\frac{n - 1}{2} x)}{\sin (\frac{x}{2})}$ $forx = \frac{2 π}{3}$ $S = \cos (\frac{n π}{3}) . \frac{\sin ((n - 1) \frac{π}{3})}{\sin (\frac{π}{3})}$ $= \frac{2}{\sqrt{3}} . \cos (\frac{n π}{3}) \sin (\frac{n - 1}{3} π)$ $\sin (\frac{n - 1}{3} π) = \sin (\frac{n π}{3}) \frac{1}{2} - \frac{\sqrt{3}}{2} \cos (n \frac{π}{3})$ $\cos (\frac{n π}{3}), \sin (\frac{n π}{3})$ ${(e^{i \frac{π}{3}})}^{n} = f (n)$ $We worck modd 6$ $n \equiv 0 (6), f (n) = 1, n \equiv 1 (6), f (n) = \frac{1}{2} + \frac{i \sqrt{3}}{2}$ $n \equiv 2 (6), f (n) = - \frac{1}{2} + \frac{i \sqrt{3}}{2}$ $n \equiv 3 (6), f (n) = - 1$ $n \equiv 4 (6), f (n) = - \frac{1}{2} - \frac{i \sqrt{3}}{2}$ $n \equiv 5 (6), f (n) = \frac{1}{2} - i \frac{\sqrt{3}}{2}$ $S = \frac{\cos (\frac{n π}{3}) \sin (\frac{n π}{3})}{\sqrt{3}} - \cos^{2} (\frac{n π}{3})$ $put different case n modd 6 get close forme add \frac{4 n - 1}{3} ge final answer$
	Commented by Raxreedoroid last updated on 01/Dec/19

	$I dont know why but I got a different solution$ $I got$ $\frac{2 \sqrt{3} c o s (\frac{2 π}{3} n - \frac{5 π}{6}) + 12 x - 6}{9}$
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