Answer:
The most correct option for the recursive expression of the geometric sequence is;
4. t₁ = 7 and tₙ = 2·tₙ₋₁, for n > 2
Step-by-step explanation:
The general form for the nth term of a geometric sequence, aₙ is given as follows;
aₙ = a₁·r⁽ⁿ⁻¹⁾
Where;
a₁ = The first term
r = The common ratio
n = The number of terms
The given geometric sequence is 7, 14, 28, 56, 112
The common ratio, r = 14/7 = 25/14 = 56/58 = 112/56 = 2
r = 2
Let, 't₁', represent the first term of the geometric sequence
Therefore, the nth term of the geometric sequence is presented as follows;
tₙ = t₁·r⁽ⁿ⁻¹⁾ = t₁·2⁽ⁿ⁻¹⁾
tₙ = t₁·2⁽ⁿ⁻¹⁾ = 2·t₁2⁽ⁿ⁻²⁾ = 2·tₙ₋₁
∴ tₙ = 2·tₙ₋₁, for n ≥ 2
Therefore, we have;
t₁ = 7 and tₙ = 2·tₙ₋₁, for n ≥ 2.
