For positive integers k=1,2,3,....n,, let `S_k` denotes the area of `triangleAOB_k` such that `angleAOB_k=(kpi)/(2n)`, OA=1 and `OB_k=k` The value of the `lim_(n->oo)1/n^2sum_(k-1)^nS_k` is

Answer

0

Explanation

`OB_(k)=k,/_AOB_(k)=(kpi)/(2n)`
`S_(k)=1/2k"sin"(kpi)/(2n)` (using `Delta=1/2 absintheta`)
`:.L=lim_(nrarroo)1/(n^(2))sum_(k=1)^(n)S_(k)=k/(2n^(2))sum_(k=1)^(oo)"sin"(kpi)/(2n)`
`=1/(2n)sum_(k=1)^(oo)k/n"sin"(kpi)/(2n)=1/2int_(0)^(1)x."sin"(pix)/2dx`
`=1/2[(-2)/(pi)xx"cos"(pix)/2|_(0)^(1)+2/(pi)int_(0)^(1)"cos"(pix)/2dx]`
`=1/2[0+2/(pi).2/(pi)("sin"(pix)/2)_(0)^(1)]=2/(pi)`