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n-2-n-3-4-n-cos-180-n-




Question Number 167003 by cortano1 last updated on 04/Mar/22
     Σ_(n=2) ^(n=∞) ((3/4))^n cos (180°n)= ?
$$\:\:\:\:\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{n}=\infty} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{n}} \mathrm{cos}\:\left(\mathrm{180}°\mathrm{n}\right)=\:? \\ $$
Answered by greogoury55 last updated on 04/Mar/22
 cos (180°n)=(−1)^n    Σ_(n=2) ^(n=∞) ((3/4))^n (−1)^n = Σ_(n=2) ^∞ (−(3/4))^n    = (−(3/4))^2 +(−(3/4))^3 +...   = ((9/(16))/(1−(−(3/4))))=((9/(16))/(7/4)) =(9/(16))×(4/7)=(9/(28))
$$\:\mathrm{cos}\:\left(\mathrm{180}°{n}\right)=\left(−\mathrm{1}\right)^{{n}} \\ $$$$\:\underset{{n}=\mathrm{2}} {\overset{{n}=\infty} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \left(−\mathrm{1}\right)^{{n}} =\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \\ $$$$\:=\:\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} +… \\ $$$$\:=\:\frac{\frac{\mathrm{9}}{\mathrm{16}}}{\mathrm{1}−\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)}=\frac{\frac{\mathrm{9}}{\mathrm{16}}}{\frac{\mathrm{7}}{\mathrm{4}}}\:=\frac{\mathrm{9}}{\mathrm{16}}×\frac{\mathrm{4}}{\mathrm{7}}=\frac{\mathrm{9}}{\mathrm{28}}\: \\ $$

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