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Question Number 56962 by turbo msup by abdo last updated on 27/Mar/19

find S_n =Σ_(k=0) ^n k^2 C_n ^k cos(2kx) interms of n.

$${find}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{cos}\left(\mathrm{2}{kx}\right) \\ $$$${interms}\:{of}\:{n}. \\ $$

Commented by maxmathsup by imad last updated on 29/Mar/19

we have S_n =Re( Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) )=Re(W_n ) let p(x)=Σ_(k=0) ^n C_n ^k e^(i2kx) =Σ_(k=0) ^n C_n ^k (e^(i2x) )^k =(1+e^(i2x) )^n but p^′ (x)=2i Σ_(k=0) ^n k C_n ^k e^(i2kx) and p^(′′) (x)=−4 Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) ⇒ Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) =−(1/4)p^((2)) (x) we have p(x)=(e^(i2x) +1)^n ⇒ p^′ (x)=2in(e^(i2x) +1)^(n−1) and p^((2)) (x)=−4n(n−1)(e^(i2x) +1)^(n−2) but (e^(i2x) +1)^(n−2) =(cos(2x)+isin(2x)+1)^(n−2) =(2cos^2 (x)+2isinx cosx)^(n−2) =(2cosx)^(n−2) (e^(ix) )^(n−2) =(2cosx)^(n−2) e^(i(n−2)x) =(2cosx)^(n−2) {cos(n−2)x +isin(n−2)x} ⇒ Σ_(k=0) ^n k^2 C_n ^k e^(i2kx) =−(1/4)(−4n(n−1)) (2cosx)^(n−2) {cos(n−2)x +isin(n−2)x} =n(n−1)2^(n−2) cos^(n−2) x{cos(n−2)x +isin(n−2)x} ⇒ S_n =n(n−1)2^(n−2) cos^(n−2) (x) cos((n−2)x) .

$${we}\:{have}\:{S}_{{n}} ={Re}\left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} {C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \right)={Re}\left({W}_{{n}} \right) \\ $$$${let}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({e}^{{i}\mathrm{2}{x}} \right)^{{k}} \:=\left(\mathrm{1}+{e}^{{i}\mathrm{2}{x}} \right)^{{n}} \: \\ $$$${but}\:{p}^{'} \left({x}\right)=\mathrm{2}{i}\:\sum_{{k}=\mathrm{0}} ^{{n}} {k}\:{C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \:\:{and}\:\:{p}^{''} \left({x}\right)=−\mathrm{4}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \:=−\frac{\mathrm{1}}{\mathrm{4}}{p}^{\left(\mathrm{2}\right)} \left({x}\right)\:\:{we}\:{have}\:{p}\left({x}\right)=\left({e}^{{i}\mathrm{2}{x}} \:+\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${p}^{'} \left({x}\right)=\mathrm{2}{in}\left({e}^{{i}\mathrm{2}{x}} \:+\mathrm{1}\right)^{{n}−\mathrm{1}} \:{and}\:{p}^{\left(\mathrm{2}\right)} \left({x}\right)=−\mathrm{4}{n}\left({n}−\mathrm{1}\right)\left({e}^{{i}\mathrm{2}{x}} \:+\mathrm{1}\right)^{{n}−\mathrm{2}} \\ $$$${but}\:\left({e}^{{i}\mathrm{2}{x}} \:+\mathrm{1}\right)^{{n}−\mathrm{2}} \:=\left({cos}\left(\mathrm{2}{x}\right)+{isin}\left(\mathrm{2}{x}\right)+\mathrm{1}\right)^{{n}−\mathrm{2}} \\ $$$$=\left(\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{2}{isinx}\:{cosx}\right)^{{n}−\mathrm{2}} =\left(\mathrm{2}{cosx}\right)^{{n}−\mathrm{2}} \left({e}^{{ix}} \right)^{{n}−\mathrm{2}} \:=\left(\mathrm{2}{cosx}\right)^{{n}−\mathrm{2}} \:{e}^{{i}\left({n}−\mathrm{2}\right){x}} \: \\ $$$$=\left(\mathrm{2}{cosx}\right)^{{n}−\mathrm{2}} \left\{{cos}\left({n}−\mathrm{2}\right){x}\:+{isin}\left({n}−\mathrm{2}\right){x}\right\}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}^{\mathrm{2}} \:{C}_{{n}} ^{{k}} \:{e}^{{i}\mathrm{2}{kx}} \:=−\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{4}{n}\left({n}−\mathrm{1}\right)\right)\:\left(\mathrm{2}{cosx}\right)^{{n}−\mathrm{2}} \left\{{cos}\left({n}−\mathrm{2}\right){x}\:+{isin}\left({n}−\mathrm{2}\right){x}\right\} \\ $$$$={n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \:{cos}^{{n}−\mathrm{2}} {x}\left\{{cos}\left({n}−\mathrm{2}\right){x}\:+{isin}\left({n}−\mathrm{2}\right){x}\right\}\:\Rightarrow \\ $$$${S}_{{n}} ={n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \:{cos}^{{n}−\mathrm{2}} \left({x}\right)\:{cos}\left(\left({n}−\mathrm{2}\right){x}\right)\:. \\ $$

Answered by Smail last updated on 27/Mar/19

S_n (x)=Σ_(k=0) ^n k^2 C_n ^k cos(2kx)=Re(Σ_(k=0) ^n k^2 ^n C_k e^(2ikx) ) z′′(x)=Σ_(k=0) ^n k^2 ^n C_k e^(2ikx) z′(x)=(1/(2i))Σ_(k=0) ^n k^n C_k e^(2ikx) +c z(x)=−(1/4)Σ_(k=0) ^n ^n C_k e^(2ikx) +cx+a =((−1)/4)(1+e^(2ix) )^n +cx+a z′(x)=(1/(2i))×ne^(2ix) (1+e^(2ix) )^(n−1) +c z′′(x)=n(e^(2ix) (1+e^(2ix) )^(n−1) +(n−1)e^(4ix) (1+e^(2ix) )^(n−2) ) =ne^(2ix) (1+e^(2ix) )^(n−2) (1+e^(2ix) +(n−1)e^(2ix) ) =ne^(2ix) (1+e^(2ix) )^(n−2) (1+ne^(2ix) ) =ne^(2ix) (1+cos(2x)+isin(2x))^(n−2) (1+ne^(2ix) ) =ne^(2ix) (2cos^2 x+2isinxcosx)^(n−2) (1+ne^(2ix) ) =ne^(2ix) 2^(n−2) cos^(n−2) (x)(e^(ix) )^(n−2) (1+ne^(2ix) ) =2^(n−2) ncos^(n−2) (x)e^(inx) (1+ne^(2ix) ) =2^(n−2) ncos^(n−2) x(e^(inx) +ne^(i(n+2)x) ) =2^(n−2) ncos^(n−2) x(cosnx+isin(nx)+ncos((n+2)x)+isin((n+2)x)) S_n =2^(n−2) ncos^(n−2) x(cosnx+ncos(n+2)x)

$${S}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} {C}_{{n}} ^{{k}} {cos}\left(\mathrm{2}{kx}\right)={Re}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:^{{n}} {C}_{{k}} {e}^{\mathrm{2}{ikx}} \right) \\ $$$${z}''\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:^{{n}} {C}_{{k}} {e}^{\mathrm{2}{ikx}} \\ $$$${z}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\:^{{n}} {C}_{{k}} {e}^{\mathrm{2}{ikx}} +{c} \\ $$$${z}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{k}} {e}^{\mathrm{2}{ikx}} +{cx}+{a} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}} +{cx}+{a} \\ $$$${z}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}×{ne}^{\mathrm{2}{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{1}} +{c} \\ $$$${z}''\left({x}\right)={n}\left({e}^{\mathrm{2}{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{1}} +\left({n}−\mathrm{1}\right){e}^{\mathrm{4}{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{2}} \right) \\ $$$$={ne}^{\mathrm{2}{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{2}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} +\left({n}−\mathrm{1}\right){e}^{\mathrm{2}{ix}} \right) \\ $$$$={ne}^{\mathrm{2}{ix}} \left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)^{{n}−\mathrm{2}} \left(\mathrm{1}+{ne}^{\mathrm{2}{ix}} \right) \\ $$$$={ne}^{\mathrm{2}{ix}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)+{isin}\left(\mathrm{2}{x}\right)\right)^{{n}−\mathrm{2}} \left(\mathrm{1}+{ne}^{\mathrm{2}{ix}} \right) \\ $$$$={ne}^{\mathrm{2}{ix}} \left(\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{2}{isinxcosx}\right)^{{n}−\mathrm{2}} \left(\mathrm{1}+{ne}^{\mathrm{2}{ix}} \right) \\ $$$$={ne}^{\mathrm{2}{ix}} \mathrm{2}^{{n}−\mathrm{2}} {cos}^{{n}−\mathrm{2}} \left({x}\right)\left({e}^{{ix}} \right)^{{n}−\mathrm{2}} \left(\mathrm{1}+{ne}^{\mathrm{2}{ix}} \right) \\ $$$$=\mathrm{2}^{{n}−\mathrm{2}} {ncos}^{{n}−\mathrm{2}} \left({x}\right){e}^{{inx}} \left(\mathrm{1}+{ne}^{\mathrm{2}{ix}} \right) \\ $$$$=\mathrm{2}^{{n}−\mathrm{2}} {ncos}^{{n}−\mathrm{2}} {x}\left({e}^{{inx}} +{ne}^{{i}\left({n}+\mathrm{2}\right){x}} \right) \\ $$$$=\mathrm{2}^{{n}−\mathrm{2}} {ncos}^{{n}−\mathrm{2}} {x}\left({cosnx}+{isin}\left({nx}\right)+{ncos}\left(\left({n}+\mathrm{2}\right){x}\right)+{isin}\left(\left({n}+\mathrm{2}\right){x}\right)\right) \\ $$$${S}_{{n}} =\mathrm{2}^{{n}−\mathrm{2}} {ncos}^{{n}−\mathrm{2}} {x}\left({cosnx}+{ncos}\left({n}+\mathrm{2}\right){x}\right) \\ $$

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