`f(n) = cot^2 (pi/n) + cot^2\ (2 pi)/n +...............+ cot^2\ ((n-1) pi)/n, ( n>1, n in N)` then `lim_(n rarr oo) f(n)/n^2` is equal to (A) `1/2` (B) `1/3` (C) `2/3` (D) `1`

Options

A. `1/2`
B. `1/3`
C. `2/3`
D. 1

Answer

B

Explanation

Let `x=-icot(kpi)/`n, where `k in {1,2,3,…..,(n-1)}`.
Then, `x/1=((cos (kpi)/n)/(isin(kpi)/n))`
`rArr (x+1)/(x-1)=(coskpi/n+isinkpi/n)/(cos(kpi)/n-isin(kpi)/n)`
`rArr (x+1)/(x-1)=(cos(kpi)/n+isin(kpi)/n)/(cos(kpi)/n-isin(kpi)/n)`
`rArr ((x+1)/(x-1))^(n)=cos(2kpi)/n+isin(2kpi)/n`
`rArr (x+1)^(n)-(x-1)^(n)=0`
`rArr .^(n)C_(1)x^(n-1)+^(n)C_(3)x^(n-3)+^(n)C_(5)x^(n-5)+.....=0`
Thus, `-icot(kpi)/n, k=1,2,3,.....,(n-1)` are roots of the above equation.
Sum of roots =0
`rArr sum_(k=1)^(n-1)(-cot(kpi)/n)=0`
Product of roots taken two at a time `=.^(n)C_(3)/.^(n)C_(5)`
Now, `f(n)=cot^(2)pi/4+cot^(2)(2pi)/n+....+cot^(2)((n-1)pi)/(n)`
`=sum_(k=1)^(n-1){-icot(kpi)/n}^(2)` 2 sum sum `(-icot(pi)/n) xx (-i cot(qpi)/n)`
`=0+2(.^(n)C_(3))/(.^(n)C_(1))=((n-1)(n-2))/(3)`
Hence, `"lim"_(n to infty)(f(n))/(n^(2))="lim"(n to infty)((n-1)(n-2))/(3n^(2))=1/3`