If F(k)=(1 + s i n (π /2 k))(1 + sin (k - 1) (π /2 k)) (1 + sin (2 k + 1) (π /2 k))(1 + sin â¡ (3 k - 1) (π /2 k)) , then the value of F(1)+F(2)+F(3) is equal to

Question

If F(k)=(1 + s i n (π /2 k))(1 + sin (k - 1) (π /2 k)) (1 + sin (2 k + 1) (π /2 k))(1 + sin â¡ (3 k - 1) (π /2 k)) , then the value of F(1)+F(2)+F(3) is equal to (A) (3/16) (B) (1/4) (C) (5/16) (D) (7/16)

Tardigrade Admin · Accepted Answer

F(1)=(1 + s i n (π /2))(1 + s i n 0)(1 + s i n (3 π /2))(1 + sin π ) =2× 1× 0× 1=0 F(2)=(1 + s i n (π /4))(1 + s i n (π /4))(1 + s i n (5 π /4))(1 + s i n (5 π /4)) =(1 + (1/√2))2(1 - (1/√2))2=(1 - (1/2))2=(1/4) F(3)=(1 + s i n (π /6))(1 + s i n (π /3))(1 + s i n (7 π /6))(1 + s i n (4 π /3)) =(1 + (1/2))(1 + (√3/2))(1 - (1/2))(1 - (√3/2)) =(1 - (1/4))(1 - (3/4))=(3/4)× (1/4)=(3/16) ⇒ F(1)+F(2)+F(3)=0+(1/4)+(3/16)=(7/16)

Q. If $F (k) = (1 + s in \frac{π}{2 k}) (1 + s in (k - 1) \frac{π}{2 k})$ $(1 + s in (2 k + 1) \frac{π}{2 k}) (1 + s in (3 k - 1) \frac{π}{2 k})$ , then the value of $F (1) + F (2) + F (3)$ is equal to

$\frac{3}{16}$

$\frac{1}{4}$

$\frac{5}{16}$

$\frac{7}{16}$

Q. If F(k)=(1+sin2kπ​)(1+sin(k−1)2kπ​) (1+sin(2k+1)2kπ​)(1+sin⁡(3k−1)2kπ​) , then the value of F(1)+F(2)+F(3) is equal to

163​

41​

165​

167​

Q. If $F (k) = (1 + s in \frac{π}{2 k}) (1 + s in (k - 1) \frac{π}{2 k})$ $(1 + s in (2 k + 1) \frac{π}{2 k}) (1 + s in (3 k - 1) \frac{π}{2 k})$ , then the value of $F (1) + F (2) + F (3)$ is equal to

$\frac{3}{16}$

$\frac{1}{4}$

$\frac{5}{16}$

$\frac{7}{16}$