If ∑i=12n sin−1xi=nπ, then ∑i=12n xi is equal to
n
2n
n(n+1)2
none of these
We have, ∑i=12n sin−1xi=nπ
⇒ sin−1x1+sin−1x2+…+sin−1x2n=nπ
Let sin−1x1−α1,sin−1x2−α2,…,sin−1xn−αn
∴α1+α2+…+α2n=nπ …….. (1)
⇒x1=sinα1,x2=sinα2,…,x2n=sinα2n∴∑i=12n xi=x1+x2+…+x2n =sinα1+sinα2+…+sinα2n
Clearly from (1), α1=π2,∀i=1,2,…,2n
∴∑i=12n xi=1+1+…+1⏟2n times =2n