`N= (3+1) (3^2 + 1) (3^4 + 1) (3^8 + 1) .......(3^64 + 1).` If `N` can be simplified as `(3^a-1)/2` then find the value of `a` :

If 3^(2n-1)= (1)/(27^(n-3)) , then the value of n is

If 3^(2n - 1) = 1/(27^(n - 3)) , then the value of n is

Let P_ (n) = (2 ^ (3) -1) / (2 ^ (3) +1) * (3 ^ (3) -1) / (3 ^ (3) +1) * (4 ^ ( 3) -1) / (4 ^ (3) +1) ...... (n ^ (3) -1) / (n ^ (3) +1) Prove that lim_ (n rarr oo) P_ ( n) = (2) / (3)

x=(3(1)/(4)+1)(3^((1)/(2))+1)(3+1)(3^(2)+1)(3^(4)+1)(3^(8)+1), then x can be written as (3^(a)-1)/(3^(b)-1) (where a is a multiple of 1/5 such that a,(1)/(b)in N), then (a.b) is equal to

Sum of the series 1+(1+2)+(1+2+3)+(1+2+3+4)+... to n terms be 1/m n(n+1)[(2n+1)/k+1] then find the value of k+m

Find the value of (3^n × 3^(2n + 1))/(3^(2n) × 3^(n - 1))

Simplify. (n + 2) / (n!) - (3n + 1) / ((n + 1)!)