If `N=2^(n-1).(2^n-1)` where `2^n-1` is a prime, then the sum of the all divisors of N is

Given that N = 2^(n)(2^(n+1) -1) and 2^(n+1) - 1 is a prime number, which of the following is true, where n is a natural number.

Given that N = 2^(n)(2^(n+1) -1) and 2^(n+1) - 1 is a prime number, which of the following is true, where n is a natural number.

N=2^(n-1)(2^(n)-1) and (2^(n)-1) is a prime number.d_(1)

P(n) : 2^(2^n) + 1 is a prime number . For n = ………., it is not true .

For each n, (2^((2^n)+1))) is a prime n in N

[ If 2 is the sum of infinity of a G.P.,whose first clement is 1 ,then the sum of the first n terms is [ 1) (2^(n)-1)/(2^(n)), 2) (2^(n)-1)/(2^(n-1)), 3) (2^(n-1)-2)/(2), 4) (2^(n-1)-1)/(2^(n))]]

If ngt3, then ((n-1),(1))(a-1)^(2)-((n-1),(2))(a-2)^(2)+((n-1),(3))(a-3)^(2)-...+(-1)^(n-2)((n-1),(n-1))(a-n+1)^(2)=k then number of positvie divisors of k, (given 'a' is a prime number)

Let f(n) = [(1)/(2) + (n)/(100)] where [n] denotes the integral part of n. Then the value of sum_(n=1)^(100) f(n) is