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Question Number 69572 by Ajao yinka last updated on 25/Sep/19

	Answered by mind is power last updated on 25/Sep/19
	$Let S_0 =Σ_(k=0) ^n (_(5k) ^(5n) ) ,S_1 =Σ_(k=0) ^(n−1) (_(5k+1) ^(5n) ),S_2 =Σ_(k=0) ^(n−1) (_(5k+2) ^(5n) ),S_3 =Σ_(k=0) ^(n−1) (_(5k+3) ^(5n) ),S_4 =Σ_(k=0) ^(n−1) (_(5k+4) ^(5n) ) ⇒S_0 +S_1 +S_2 +S_3 +S_4 =Σ_(k=0) ^(5n) (_k ^(5n) )=2^(5n) ......1 Let f=(1+e^((2iπ)/5) )^(5n) =e^(inπ) (e^((iπ)/5) +e^((−iπ)/5) )^(5n) =(−1)^n ∗2^(5n) cos^(5n) ((π/5)) f=(1+e^((2iπ)/5) )^(5n) =Σ_(k=0) ^(5n) (_k ^(5n) )e^((2ikπ)/5) =Σ_(k=0) ^n (_(5k) ^(5n) )+Σ_(k=0) ^(n−1) (_(5k+1) ^(5n) )e^((2iπ)/5) +Σ_(k=0) ^(n−1) (_(5k+2) ^(5n) )e^((4iπ)/5) +Σ_(k=0) ^(n−1) (_(5k+3) ^(5n) )e^((6iπ)/5) +Σ_(k=0) ^(n−1) (_(5k+4) ^(5n) )e^((8iπ)/5) =S_0 +e^((2iπ)/5) S_1 +e^((4iπ)/5) S_2 +e^((−4iπ)/5) S_3 +e^((−2iπ)/5) S_4 =(−1)^n 2^(5n) cos^(5n) ((π/5)) We Tack real and Im part⇒ S_0 +cos(((2π)/5))(S_1 +S_4 )+cos(((4π)/5))(S_2 +S_3 )=(−1)^n 2^(5n) ×cos^(5n) ((π/5))...2 sin(((2π)/5))(S_1 −S_4 )+sin(((4π)/5))(S_2 −S_3 )=0...3 Now let Evaluat (1+e^((4iπ)/5) )^(5n) =(e^((2iπ)/5) (e^((2iπ)/5) +e^((−2iπ)/5) ))^(5n) =e^(2inπ) (2cos(((2π)/5)))^(5n) =2^(5n) cos^(5n) (((2π)/5)) (1+e^((4iπ)/5) )^(5n) =S_0 +e^((4iπ)/5) S_1 +e^((−2iπ)/5) S_2 +e^((2iπ)/5) S_3 +e^(−((4iπ)/5)) S_4 =2^(5n) cos^(5n) (((2π)/5)) ⇒ S_0 +cos(((4π)/5))(S_1 +S_4 )+cos(((2π)/5))(S_2 +S_3 )=2^(5n) cos^(5n) (((2π)/5))...4 and (S_1 −S_4 )sin(((4π)/5))+(S_3 −S_2 )sin(((2π)/5))=0 ......5 3&5⇔ (((sin(((2π)/5)) sin(((4π)/5)))),((sin(((4π)/5)) −sin(((2π)/5)))) ) ((((S_1 −S_4 ))),(((S_2 −S_3 ))) )= ((0),(0) ) Det [((sin(((2π)/5)) sin(((4π)/5)))),((sin(((4π)/5)) −sin(((2π)/5)))) ]≠0⇒S_1 =S_4 &S_2 =S_3 Substitution S_4 by S_1 and S_3 byS_2 Our equation Becom { ((S_0 +2S_1 +2S_2 =2^(5n) ....a)),((S_0 +2S_1 cos(((2π)/5))+2S_2 cos(((4π)/5))=(−1)^n 2^(5n) cos^(5n) ((π/5))..b)) :} and S_0 +2S_1 cos(((4π)/5))+2S_2 cos (((2π)/5))=2^(5n) cos^(5n) (((2π)/5))...c after See that 4cos^2 (((2π)/5))+2cos(((2π)/5))−1=0=2cos(((4π)/5))+2cos(((2π)/5))+1 ⇒a×(1/2)+b+c=(5/2)S_0 +S_1 (1+2cos(((2π)/5))+2cos(((4π)/5)))+S_2 (1+2cos(((2π)/5))+cos(((4π)/5)))=2^(5n) ((1/2)+(−1)^n cos^(5n) ((π/5))+cos^(5n) (((2π)/5))) ⇒(5/2)S_0 =2^(5n) ((1/2)+(−1)^n cos^(5n) ((π/5))+cos^(5n) (((2π)/5))) ⇒S_0 =(2^(5n) /5)(1+2((−1)^n cos^(5n) ((π/5))+cos^(5n) (((2π)/5)))) ⇔Σ_(k=0) ^(5n) (((5n)),((5k)) )=(2^(5n) /5)(1+2((−1)^n cos^(5n) ((π/5))+cos^(5n) (((2π)/5))))$
	$L e t S_{0} = \underset{k = 0}{\sum^{n}} (_{5 k}^{5 n}), S_{1} = \underset{k = 0}{\sum^{n - 1}} (_{5 k + 1}^{5 n}), S_{2} = \underset{k = 0}{\sum^{n - 1}} (_{5 k + 2}^{5 n}), S_{3} = \underset{k = 0}{\sum^{n - 1}} (_{5 k + 3}^{5 n}), S_{4} = \underset{k = 0}{\sum^{n - 1}} (_{5 k + 4}^{5 n})$ $\Rightarrow S_{0} + S_{1} + S_{2} + S_{3} + S_{4} = \underset{k = 0}{\sum^{5 n}} (_{k}^{5 n}) = 2^{5 n} . . . . . . 1$ $L e t f = {(1 + e^{\frac{2 i π}{5}})}^{5 n} = e^{i n π} {(e^{\frac{i π}{5}} + e^{\frac{- i π}{5}})}^{5 n} = {(- 1)}^{n} * 2^{5 n} {c o s}^{5 n} (\frac{π}{5})$ $f = {(1 + e^{\frac{2 i π}{5}})}^{5 n} = \underset{k = 0}{\sum^{5 n}} (_{k}^{5 n}) e^{\frac{2 i k π}{5}} = \underset{k = 0}{\sum^{n}} (_{5 k}^{5 n}) + \underset{k = 0}{\sum^{n - 1}} (_{5 k + 1}^{5 n}) e^{\frac{2 i π}{5}} + \underset{k = 0}{\sum^{n - 1}} (_{5 k + 2}^{5 n}) e^{\frac{4 i π}{5}} + \underset{k = 0}{\sum^{n - 1}} (_{5 k + 3}^{5 n}) e^{\frac{6 i π}{5}} + \underset{k = 0}{\sum^{n - 1}} (_{5 k + 4}^{5 n}) e^{\frac{8 i π}{5}}$ $= S_{0} + e^{\frac{2 i π}{5}} S_{1} + e^{\frac{4 i π}{5}} S_{2} + e^{\frac{- 4 i π}{5}} S_{3} + e^{\frac{- 2 i π}{5}} S_{4} = {(- 1)}^{n} 2^{5 n} {c o s}^{5 n} (\frac{π}{5})$ $W e T a c k r e a l a n d I m p a r t \Rightarrow$ $S_{0} + c o s (\frac{2 π}{5}) (S_{1} + S_{4}) + c o s (\frac{4 π}{5}) (S_{2} + S_{3}) = {(- 1)}^{n} 2^{5 n} \times {c o s}^{5 n} (\frac{π}{5}) . . . 2$ $s i n (\frac{2 π}{5}) (S_{1} - S_{4}) + s i n (\frac{4 π}{5}) (S_{2} - S_{3}) = 0 . . . 3$ $N o w l e t E v a l u a t {(1 + e^{\frac{4 i π}{5}})}^{5 n} = {(e^{\frac{2 i π}{5}} (e^{\frac{2 i π}{5}} + e^{\frac{- 2 i π}{5}}))}^{5 n} = e^{2 i n π} {(2 c o s (\frac{2 π}{5}))}^{5 n} = 2^{5 n} {c o s}^{5 n} (\frac{2 π}{5})$ ${(1 + e^{\frac{4 i π}{5}})}^{5 n} = S_{0} + e^{\frac{4 i π}{5}} S_{1} + e^{\frac{- 2 i π}{5}} S_{2} + e^{\frac{2 i π}{5}} S_{3} + e^{- \frac{4 i π}{5}} S_{4} = 2^{5 n} {c o s}^{5 n} (\frac{2 π}{5})$ $\Rightarrow$ $S_{0} + c o s (\frac{4 π}{5}) (S_{1} + S_{4}) + c o s (\frac{2 π}{5}) (S_{2} + S_{3}) = 2^{5 n} \cos^{5 n} (\frac{2 π}{5}) . . . 4$ $a n d (S_{1} - S_{4}) s i n (\frac{4 π}{5}) + (S_{3} - S_{2}) s i n (\frac{2 π}{5}) = 0 . . . . . . 5$ $3 & 5 \Leftrightarrow$ $(\begin{matrix} s i n (\frac{2 π}{5}) s i n (\frac{4 π}{5}) \\ s i n (\frac{4 π}{5}) - s i n (\frac{2 π}{5}) \end{matrix}) (\begin{matrix} (S_{1} - S_{4}) \\ (S_{2} - S_{3}) \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ $D e t [\begin{matrix} s i n (\frac{2 π}{5}) s i n (\frac{4 π}{5}) \\ s i n (\frac{4 π}{5}) - s i n (\frac{2 π}{5}) \end{matrix}] \neq 0 \Rightarrow S_{1} = S_{4} & S_{2} = S_{3}$ $S u b s t i t u t i o n S_{4} b y S_{1} a n d S_{3} {b y S}_{2} O u r e q u a t i o n B e c o m$ ${\begin{cases} S_{0} + 2 S_{1} + 2 S_{2} = 2^{5 n} . . . . a \\ S_{0} + 2 S_{1} c o s (\frac{2 π}{5}) + 2 S_{2} c o s (\frac{4 π}{5}) = {(- 1)}^{n} 2^{5 n} \cos^{5 n} (\frac{π}{5}) . . b \end{cases}$ $a n d S_{0} + 2 S_{1} c o s (\frac{4 π}{5}) + 2 S_{2} \cos (\frac{2 π}{5}) = 2^{5 n} {c o s}^{5 n} (\frac{2 π}{5}) . . . c$ $a f t e r S e e t h a t 4 {c o s}^{2} (\frac{2 π}{5}) + 2 c o s (\frac{2 π}{5}) - 1 = 0 = 2 c o s (\frac{4 π}{5}) + 2 c o s (\frac{2 π}{5}) + 1$ $\Rightarrow a \times \frac{1}{2} + b + c = \frac{5}{2} S_{0} + S_{1} (1 + 2 c o s (\frac{2 π}{5}) + 2 c o s (\frac{4 π}{5})) + S_{2} (1 + 2 c o s (\frac{2 π}{5}) + c o s (\frac{4 π}{5})) = 2^{5 n} (\frac{1}{2} + {(- 1)}^{n} {c o s}^{5 n} (\frac{π}{5}) + {c o s}^{5 n} (\frac{2 π}{5}))$ $\Rightarrow \frac{5}{2} S_{0} = 2^{5 n} (\frac{1}{2} + {(- 1)}^{n} {c o s}^{5 n} (\frac{π}{5}) + {c o s}^{5 n} (\frac{2 π}{5}))$ $\Rightarrow S_{0} = \frac{2^{5 n}}{5} (1 + 2 ({(- 1)}^{n} {c o s}^{5 n} (\frac{π}{5}) + {c o s}^{5 n} (\frac{2 π}{5})))$ $\Leftrightarrow \underset{k = 0}{\sum^{5 n}} (\begin{matrix} 5 n \\ 5 k \end{matrix}) = \frac{2^{5 n}}{5} (1 + 2 ({(- 1)}^{n} \cos^{5 n} (\frac{π}{5}) + \cos^{5 n} (\frac{2 π}{5})))$
	Commented by Ajao yinka last updated on 26/Sep/19

	$n i c e o n e$
	Commented by otchereabdullai@gmail.com last updated on 29/Sep/19

	$powerful$
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