How does the free fall of a body from a state of rest increase the speed in a third second?

V0 = 0 m / s.
t1 = 2 s.
t2 = 3 s.
t = 1 s.
g = 9.8 m / s ^ 2.
V2 – V1 -?
The body is in a state of free fall, under the influence of only gravity. It moves with gravitational acceleration g = 9.8 m / s ^ 2.
1) According to the definition, acceleration is determined by the formula: g = (V2 – V1) / t.
V2 – V1 = g * t.
V2 – V1 = 9.8 m / s ^ 2 * 1s = 9.8 m / s.
For any 1 second of movement, the speed of the body will increase by 9.8 m / s.
2) V1 = g * t1.
V2 = g * t2.
V2 – V1 = g * t2 – g * t1 = g * (t2 – t1).
V2 – V1 = 9.8 m / s ^ 2 * (3 s – 2s) = 9.8 m / s.
Answer: for any 1 second of movement, the speed of the body will increase by 9.8 m / s.