`sum_(i=4)^7 sqrt(2i-1)" "`can be written as

The complex number (i + sqrt 3 ) in polar can be written as :

If sum_(i=1)^(2n) sin^(-1) x_i=npi , then sum_(i=1)^(2n) x_i equals :

If sum_(i=1)^(2n) sin^(-1) x_i =npi , then sum_(i=1)^(2n) x_i is equal to :

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)

The sum 1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!) can be written in the form 2^a/(b!) Find a and b.

If sum_(i=1)^(2n)sin^(-1)x_(i)=n pi then find the value of sum_(i=1)^(2n)x_(i)

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 backslash 'n' observations (x_(1)) is: (1)2sqrt(3)(2)sqrt(3)(3)sqrt(5)(4)3sqrt(2)

If sum_(i=1)^(7) i^(2)x_(i) = 1 and sum_(i=1)^(7)(i+1)^(2) x_(i) = 12 and sum_(i=1)^(7)(i+2)^(2)x_(i) = 123 then find the value of sum_(i=1)^(7)(i+3)^(2)x_(i)"____"

If sum_(i=1)^(7) i^(2)x_(i) = 1 and sum_(i=1)^(7)(i+1)^(2) x_(i) = 12 and sum_(i=1)^(7)(i+2)^(2)x_(i) = 123 then find the value of sum_(i=1)^(7)(i+3)^(2)x_(i)"____"