If the property of the elements of the set B is 1+3+9+..+3^(n+1)=1093 what are the natural elements of the set.

Let n be the elements of the set
B.

To determine the elements of the set, we'll have to
solve the sum of the consecutive terms of a geometric progresison, that represents the
property of the elements of the set.

The number of terms of
the geometric progression is n+2. The common ratio of the geometric progression is q =
3.

The sum of n+2 terms of the geometric progression
is:

S = b1*(q^(n+2) -
1)/(q-1)

Foe b1 = 1 and q = 3, we'll
get:

S = [3^(n+2) - 1]/(3-1)

S
= [3^(n+2) - 1]/2

But S = 1093 => [3^(n+2) - 1]/2 =
1093

Therefore, we'll
have:

3^(n+2) - 1 =
2186

3^(n+2) = 2187

We'll
create matching bases since 2187 = 3^7:

3^(n+2) =
3^7

Since the bases are matching, we'll apply one to one
property:

n + 2 = 7

n =
5

The natural number that has the given
property of the set B is n = 5.