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))))](https://www.tinkutara.com/question/Q69592.png)
$${Let}\:{S}_{\mathrm{0}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{5}{k}} ^{\mathrm{5}{n}} \right)\:\:\:,{S}_{\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{1}} ^{\mathrm{5}{n}} \right),{S}_{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{2}} ^{\mathrm{5}{n}} \right),{S}_{\mathrm{3}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{3}} ^{\mathrm{5}{n}} \right),{S}_{\mathrm{4}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{4}} ^{\mathrm{5}{n}} \right) \\ $$$$\Rightarrow{S}_{\mathrm{0}} +{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}{n}} {\sum}}\left(_{{k}} ^{\mathrm{5}{n}} \right)=\mathrm{2}^{\mathrm{5}{n}} ……\mathrm{1} \\ $$$${Let}\:\:{f}=\left(\mathrm{1}+{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \right)^{\mathrm{5}{n}} ={e}^{{in}\pi} \left({e}^{\frac{{i}\pi}{\mathrm{5}}} +{e}^{\frac{−{i}\pi}{\mathrm{5}}} \right)^{\mathrm{5}{n}} =\left(−\mathrm{1}\right)^{{n}} \ast\mathrm{2}^{\mathrm{5}{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right) \\ $$$${f}=\left(\mathrm{1}+{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \right)^{\mathrm{5}{n}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}{n}} {\sum}}\left(_{{k}} ^{\mathrm{5}{n}} \right){e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{5}}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{5}{k}} ^{\mathrm{5}{n}} \right)+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{1}} ^{\mathrm{5}{n}} \right){e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{2}} ^{\mathrm{5}{n}} \right){e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{3}} ^{\mathrm{5}{n}} \right){e}^{\frac{\mathrm{6}{i}\pi}{\mathrm{5}}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(_{\mathrm{5}{k}+\mathrm{4}} ^{\mathrm{5}{n}} \right){e}^{\frac{\mathrm{8}{i}\pi}{\mathrm{5}}} \\ $$$$={S}_{\mathrm{0}} +{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{1}} +{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{2}} +{e}^{\frac{−\mathrm{4}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{3}} +{e}^{\frac{−\mathrm{2}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{4}} =\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{\mathrm{5}{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right) \\ $$$${We}\:{Tack}\:{real}\:{and}\:{Im}\:{part}\Rightarrow \\ $$$$\:\:{S}_{\mathrm{0}} +{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{1}} +{S}_{\mathrm{4}} \right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{2}} +{S}_{\mathrm{3}} \right)=\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{\mathrm{5}{n}} ×{cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)…\mathrm{2} \\ $$$${sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{1}} −{S}_{\mathrm{4}} \right)+{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{2}} −{S}_{\mathrm{3}} \right)=\mathrm{0}…\mathrm{3} \\ $$$${Now}\:{let}\:{Evaluat}\:\left(\mathrm{1}+{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} \right)^{\mathrm{5}{n}} =\left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} \left({e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} +{e}^{\frac{−\mathrm{2}{i}\pi}{\mathrm{5}}} \right)\right)^{\mathrm{5}{n}} ={e}^{\mathrm{2}{in}\pi} \left(\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right)^{\mathrm{5}{n}} =\mathrm{2}^{\mathrm{5}{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\left(\mathrm{1}+{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} \right)^{\mathrm{5}{n}} ={S}_{\mathrm{0}} +{e}^{\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{1}} +{e}^{\frac{−\mathrm{2}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{2}} +{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{3}} +{e}^{−\frac{\mathrm{4}{i}\pi}{\mathrm{5}}} {S}_{\mathrm{4}} =\mathrm{2}^{\mathrm{5}{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$$\Rightarrow \\ $$$${S}_{\mathrm{0}} +{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{1}} +{S}_{\mathrm{4}} \right)+{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\left({S}_{\mathrm{2}} +{S}_{\mathrm{3}} \right)=\mathrm{2}^{\mathrm{5}{n}} \mathrm{cos}\:^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)…\mathrm{4} \\ $$$${and}\:\left({S}_{\mathrm{1}} −{S}_{\mathrm{4}} \right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)+\left({S}_{\mathrm{3}} −{S}_{\mathrm{2}} \right){sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\mathrm{0}\:\:\:\:……\mathrm{5} \\ $$$$\mathrm{3\&5}\Leftrightarrow \\ $$$$\begin{pmatrix}{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:\:\:\:\:\:\:\:{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)}\\{{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:\:\:\:−{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}\end{pmatrix}\:\:\:\:\begin{pmatrix}{\left({S}_{\mathrm{1}} −{S}_{\mathrm{4}} \right)}\\{\left({S}_{\mathrm{2}} −{S}_{\mathrm{3}} \right)}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${Det}\begin{bmatrix}{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:\:\:\:\:\:\:{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)}\\{{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:\:−{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}\end{bmatrix}\neq\mathrm{0}\Rightarrow{S}_{\mathrm{1}} ={S}_{\mathrm{4}} \&{S}_{\mathrm{2}} ={S}_{\mathrm{3}} \\ $$$${Substitution}\:{S}_{\mathrm{4}} {by}\:{S}_{\mathrm{1}} {and}\:{S}_{\mathrm{3}} {byS}_{\mathrm{2}} \:{Our}\:{equation}\:{Becom} \\ $$$$\begin{cases}{{S}_{\mathrm{0}} +\mathrm{2}{S}_{\mathrm{1}} +\mathrm{2}{S}_{\mathrm{2}} =\mathrm{2}^{\mathrm{5}{n}} ….{a}}\\{{S}_{\mathrm{0}} +\mathrm{2}{S}_{\mathrm{1}} {cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{2}{S}_{\mathrm{2}} {cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)=\left(−\mathrm{1}\right)^{{n}} \mathrm{2}^{\mathrm{5}{n}} \mathrm{cos}\:^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)..{b}}\end{cases} \\ $$$${and}\:{S}_{\mathrm{0}} +\mathrm{2}{S}_{\mathrm{1}} {cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)+\mathrm{2}{S}_{\mathrm{2}} \mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\mathrm{2}^{\mathrm{5}{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)…{c} \\ $$$${after}\:{See}\:{that}\:\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)−\mathrm{1}=\mathrm{0}=\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow{a}×\frac{\mathrm{1}}{\mathrm{2}}+{b}+{c}=\frac{\mathrm{5}}{\mathrm{2}}{S}_{\mathrm{0}} +{S}_{\mathrm{1}} \left(\mathrm{1}+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\right)+{S}_{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\right)=\mathrm{2}^{\mathrm{5}{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\left(−\mathrm{1}\right)^{{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)+{cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right) \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{2}}{S}_{\mathrm{0}} =\mathrm{2}^{\mathrm{5}{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\left(−\mathrm{1}\right)^{{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)+{cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right) \\ $$$$\Rightarrow{S}_{\mathrm{0}} =\frac{\mathrm{2}^{\mathrm{5}{n}} }{\mathrm{5}}\left(\mathrm{1}+\mathrm{2}\left(\left(−\mathrm{1}\right)^{{n}} {cos}^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)+{cos}^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right)\right) \\ $$$$\Leftrightarrow\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}{n}} {\sum}}\begin{pmatrix}{\mathrm{5}{n}}\\{\mathrm{5}{k}}\end{pmatrix}=\frac{\mathrm{2}^{\mathrm{5}{n}} }{\mathrm{5}}\left(\mathrm{1}+\mathrm{2}\left(\left(−\mathrm{1}\right)^{{n}} \mathrm{cos}\:^{\mathrm{5}{n}} \left(\frac{\pi}{\mathrm{5}}\right)+\mathrm{cos}\:^{\mathrm{5}{n}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right)\right) \\ $$$$ \\ $$
Commented by Ajao yinka last updated on 26/Sep/19
$${nice}\:{one} \\ $$
Commented by otchereabdullai@gmail.com last updated on 29/Sep/19
$$\mathrm{powerful} \\ $$