Let F k - 1 - sin-2k 1 - sin k - 1-2k 1 - sin 2k - 1-2k 1 - sin 3k - 1-2k-The value of F 1 - F 2 - F 3 is equal to

The correct option is C 716 F(k)=(1+sinπ2k)(1+sin(k 1)π2k)(1+sin(2k+1)π2k)(1+sin(3k 1)π2k) F(1)+F(2)+F(3)=(2)(1)(0)+(√(2)+1√(2))^2( 1+√(2)√(2))^ ...

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Definition of Functions

Let F k = 1...

Question

Let $F (k) = (1 + sin \frac{π}{2 k}) (1 + sin (k - 1) \frac{π}{2 k}) (1 + sin (2 k + 1) \frac{π}{2 k}) (1 + sin (3 k - 1) \frac{π}{2 k})$ .
The value of $F (1) + F (2) + F (3)$ is equal to

\frac{3}{16}

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\frac{1}{4}

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\frac{5}{16}

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\frac{7}{16}

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Solution

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