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\begin{array} {l}[br]\displaystyle \cos \alpha + \cos ( \alpha + 2 \beta ) + \ldots + \cos ( \alpha + 2 n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} + e^{i(\alpha+2\beta)} + \ldots + e^{i(\alpha+2n\beta)} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} ( 1+e^{i2\beta}+\ldots+e^{i2n\beta}) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (\frac{e^{i2(n+1)\beta} - 1}{e^{i2\beta}-1} \right ) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \frac{e^{i(n+1)\beta}}{e^{i\beta}} \frac{e^{i(n+1)\beta}-e^{-i(n+1)\beta}}{e^{i\beta}-e^{-i\beta}} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i (\alpha + n\beta)} \frac{\sin(n+1)\beta}{\sin \beta} \right ) \\[br]\displaystyle = \frac{\sin (n+1)\beta \cos (\alpha + n \beta)}{\sin \beta} \end{array}
\begin{array}{l}[br]\displaystyle \cos \alpha + \binom{n}{1} \cos ( \alpha + 2\beta ) + \ldots + \binom{n}{r} \cos ( \alpha + 2r \beta ) + \ldots + \cos ( \alpha + 2n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (1+e^{i2\beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i \alpha} \left ( 2 \cos \beta e^{i \beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i( \alpha + n \beta )} 2^n cos ^n \beta \right ) \\[br]\displaystyle = 2^n \cos^n \beta \cos ( \alpha + n \beta) [br]\end{array}[br]
\begin{array}{l} \displaystyle 1+\cos \theta \sec \theta + \ldots + \cos n \theta \sec ^n \theta +\cos (n+1)\theta \sec^{n+1} \theta \\ [br]\displaystyle = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta} + \cos (n+1)\theta \sec^{n+1} \theta \\[br]= \displaystyle \frac{\sin(n+1)\theta \sec^n \theta + \sin \theta \cos (n+1) \theta \sec^{n+1}}{\sin \theta} \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \left ( \sin(n+1)\theta \cos \theta + \sin \theta \cos (n+1) \theta \right ) \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \sin(n+2) \end{array}
\displaystyle \boxed{\Pr(N=n)} = \sum_{m} \Pr(M=m,N=n) = \sum_{m=0}^{\infty} \left ( \frac{1}{4} \right )^{m+1} \left ( \frac{3}{4} \right )^n \\[br]= \left ( \frac{3}{4} \right )^n \sum_{m=0}^{\infty} \left ( \frac{1}{4} \right )^{m+1} = \left ( \frac{3}{4} \right )^n \frac{1}{3}
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