`sqrt(n^3+2) - sqrt(n^3-2)`

The sum of the series (1)/(sqrt(1)+sqrt(2))+(1)/(sqrt(2)+sqrt(3))+(1)/(sqrt(3)+sqrt(4))+ . . . . .+(1)/(sqrt(n^(2)-1)+sqrt(n^(2))) equals

lim _(x to oo) (1)/(n ^(3))(sqrt(n ^(2)+1)+2 sqrt(n ^(2) +2 ^(2))+ .... + n sqrt((n ^(2) + n ^(2)))=:

lim_(n->oo)[1/sqrt(2n-1^2) +1/sqrt(4n-2^2)+1/sqrt(6n-3^2)+...+1/n]

If A >0,\ B >0\ a n d\ A+B=\ pi/6 , then the minimum value of t a n A+t a n B is: 2-sqrt(3) b. 4-2sqrt(3) c. sqrt(3)-sqrt(2) d. 2/(sqrt(3))

Let f (n)=(4n + sqrt(4n ^(2) -1))/( sqrt(2n +1 )+sqrt(2n-1)),n in N then the remainder when f (1) + f (2) + f (3) + ..... + f (60) is divided by 9 is.

If C_r=(n !)/([r !(n-r)]), the prove that sqrt(C_1)+sqrt(C_2)+.......sqrt(C_n) lt sqrt(n(2^n-1)) ="">

Sum of 1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+1/(sqrt(11)+sqrt(14))+..to n terms= (A) n/(sqrt(3n+2)-sqrt(2)) (B) 1/3 (sqrt(2)-sqrt(3n+2) (C) n/(sqrt(3n+2)+sqrt(2)) (D) none of these

If sum_(i=1)^n(x_i+1)^2=9n and sum_(i=1)^n(x_i-1)^2=5n , then standard deviation of these 'n' observations (x_1) is: (1) 2sqrt(3) (2) sqrt(3) (3) sqrt(5) (4) 3sqrt(2)

Evaluate: ("lim")_(n rarr oo)(1/(sqrt(4n^2-1))+1/(sqrt(4n^2-2^2))++1/(sqrt(3n^2)))

In each of the following determine rational number a\ a n d\ b : (i) (sqrt(3)-1)/(sqrt(3)+1)=a-bsqrt(3) (ii) (4+\ sqrt(2))/(2+sqrt(2))=a-sqrt(b)