Q.

limt→0(11sin2t+21sin2t+⋯+n1sin2t)sin2t   is equal to               [2023]

1 n2  
2 n2+n  
3 n  
4 n(n+1)2  

Ans.

(3)

Let L=limt→0(11sin2t+21sin2t+…+n1sin2t)sin2t

⇒L=limt→0n[(1n)cosec2t+(2n)cosec2t+ …+(n-1n)cosec2t+1]sin2t

⇒L=n(1sin(0))=n(1)=n[0<kn<1,(kn)∞=0 for 1≤k<n]