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Question Number 54675 by maxmathsup by imad last updated on 09/Feb/19

$$simplify{A}_n=\sum _{k=0}^n{k}^2{C}_n^{k}cos\left(k\theta \right)and{B}_n=\sum _{k=0}^n{k}^2{C}_n^{k}sin\left(k\theta \right).$$

Commented by maxmathsup by imad last updated on 10/Feb/19

$$wehave{A}_n+{iB}_n=\sum _{k=0}^n{k}^2{C}_n^k{\left(e^{i\theta }\right)}^{k}butwehaveprovedthat$$$$\sum _{k=0}^n{k}^2{C}_n^k{z}^k=nz{(z+1)}^{n-1}+n(n-1)z^2{(z+1)}^{n-2}\Rightarrow$$$$\sum _{k=0}^n{k}^2{C}_n^k{\left(e^{i\theta }\right)}^k=n{e}^{i\theta }{(1+e^{i\theta })}^{n-1}+n(n-1)e^{2i\theta }{(e^{i\theta }+1)}^{n-2}but$$$$1+e^{i\theta }=1+cos\theta +isin\theta =2{cos}^2\left(\frac{\theta }{2}\right)+2isin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)$$$$=2cos\left(\frac{\theta }{2}\right)e^{i\frac{\theta }{2}}\Rightarrow A_n+{iB}_n=n{e}^{i\theta }{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)e^{i\frac{n\theta }{2}}$$$$+n(n-1)e^{2i\theta }{2}^{n-2}{cos}^{n-2}\left(\frac{\theta }{2}\right)e^{i\frac{(n-2)\theta }{2}}$$$$=n{2}^{n-1}{cos}^{n-1)}\left(\frac{\theta }{2}\right)e^{i(\theta +\frac{n\theta }{2})}+n(n-1)2^{n-2}{e}^{i(2\theta +\frac{(n-2)\theta }{2})}$$$$=n{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)\{cos\left(\frac{(n+2)\theta }{2}\right)+isin\left(\frac{(n+2)\theta }{2}\right)\}$$$$+n(n-1)2^{n-2}\{cos\left(\frac{(n+2)\theta }{2}\right)+isin\left(\frac{(n+2)\theta }{2}\right)$$$$=(n{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)+n(n-1)2^{n-2})cos\left(\frac{(n+2)\theta }{2}\right)$$$$+i(n{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)+n(n-1)2^{n-2})sin\left(\frac{(n+2)\theta }{2}\right)\Rightarrow$$$$A_n=(n{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)+n(n-1)2^{n-2})cos\left(\frac{(n+2)\theta }{2}\right)and$$$$B_n=(n{2}^{n-1}{cos}^{n-1}\left(\frac{\theta }{2}\right)+n(n-1)2^{n-2})sin\left(\frac{(n+2)\theta }{2}\right).$$

Answered by Smail last updated on 09/Feb/19

$$A_n+{iB}_n=S^{″}\left(\theta \right)$$$$S^{″}\left(\theta \right)=\underset{k=0}{\sum ^n}{k}^2{C}_n^k{e}^{i\theta k}$$$$S^{\prime }\left(\theta \right)=\frac{1}{i}\underset{k=0}{\sum ^n}{kC}_n^k{e}^{i\theta k}+c_1$$$$S\left(\theta \right)=-\underset{k=0}{\sum ^n}{C}_n^k{e}^{i\theta k}+\theta c_1+c_2$$$$S\left(\theta \right)=-{(1+e^{i\theta })}^n+\theta c_1+c_2$$$$S^{\prime }\left(\theta \right)=-{ine}^{i\theta }{(1+e^{i\theta })}^{n-1}+c_1$$$$S^{″}\left(\theta \right)={ne}^{i\theta }{(1+e^{i\theta })}^{n-1}+n(n-1)e^{2i\theta }{(1+e^{i\theta })}^{n-2}$$$$={ne}^{i\theta }{(1+e^{i\theta })}^{n-2}(1+e^{i\theta }+(n-1)e^{i\theta })$$$$A_n+{iB}_n={ne}^{i\theta }{(1+e^{i\theta })}^{n-2}(1+{ne}^{i\theta })$$$$={ne}^{i\theta }{(1+cos\theta +isin\theta )}^{n-2}(1+{ne}^{i\theta })$$$$={ne}^{i\theta }{(2{cos}^2\frac{\theta }{2}+2isin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right))}^{n-2}(1+{ne}^{i\theta })$$$$=2^{n-2}{ne}^{i\theta }{cos}^{n-2}\frac{\theta }{2}{(cos\frac{\theta }{2}+isin\frac{\theta }{2})}^{n-2}(1+{ne}^{i\theta })$$$$=2^{n-2}{ne}^{i\theta }{cos}^{n-2}\frac{\theta }{2}\left(e^{i\frac{n-2}{2}\theta }\right)(1+{ne}^{i\theta })$$$$=2^{n-2}{ncos}^{n-2}\left(\frac{\theta }{2}\right)e^{i\frac{n}{2}\theta }(1+{ne}^{i\theta })$$$$=2^{n-2}{ncos}^{n-2}\left(\frac{\theta }{2}\right)(e^{i\frac{n}{2}\theta }+{ne}^{i\frac{n+2}{2}\theta })$$$$=2^{n-2}{ncos}^{n-2}\left(\frac{\theta }{2}\right)(cos\frac{n\theta }{2}+isin\frac{n\theta }{2}+ncos\frac{n+2}{2}\theta +insin\frac{n+2}{2}\theta )$$$$A_n=2^{n-2}{ncos}^{n-2}\frac{\theta }{2}(cos\frac{n\theta }{2}+ncos\frac{n+2}{2}\theta )$$$$B_n=2^{n-2}{ncos}^{n-2}\frac{\theta }{2}(sin\frac{n\theta }{2}+nsin\frac{n+2}{2}\theta )$$

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