Sn=n∑a=2sin−1[√(a2+2a)−√(a−1)(a+1)a(a+1)]
Putting 1a=x,1a+1=y
Let,
L=[√(a2+2a)−√(a−1)(a+1)a(a+1)]
=xy[√(a+1)2−1−√a2−1]
=xy[√1y2−1−√1x2−1]
=x√1−y2−y√1−x2
∴Ta=sin−1(x√1−y2−y√1−x2)
=sin−1x−sin−1y =sin−11a−sin−11a+1
Substituting a=2,3,4,...,n
Sn=sin−112−sin−113 +sin−113−sin−114 +sin−114−sin−115 : : +sin−11n−sin−11n+1
limn→∞Sn=limn→∞(sin−112−sin−11n+1)
=π6+0=π6
⇒k=20