The value of `[1^2+2^2+3^2+...n^2]/n^3` is

If the variate of a distribution takes the values 1^(2), 2^(2), 3^(2),….n^(2) with frequencies n, n-1, n-2,….3,2,1 respectively, then the mean value of the distribution is

the value of 1.1!+ 2.2!+ 3.3!+ ....+ n.n!

The value of quad 1^(6)+2^(6)+3^(6)...n^(6)lim_(n rarr oo)(1^(6)+2^(6)+3^(6)...n^(6))/((1^(2)+2^(2)+3^(2)+...n^(2))(1^(3)+2^(3)+...n^(3)))

If 2^(2n-1)=(1)/(8^(n-3)), then the value of n is a -2 b.2 c.0 d.3

Let 'sigma' denotes the sum of the infinite series sum_(n=1)^(n)((n^(2)+2n+3)/(2^(n))). Compute the value of (1^(3)+2^(3)+3^(3)+......+sigma^(3))

The value of (2^(n+4)-2*2^(n))/(2.2^(n+3))+2^(-3)

The value of i^(2n)+i^(2n+1)+i^(2n+2)+i^(2n+3), where i=sqrt(-1), is

Find the value of n if 2n+1,n^(2)+n+1,3n^(2)-3n+3 are consecutive terms of an AP