Answer:
t₁ = 3
t₂ = 7
t₃ = 11
[tex]t_{(n+1)} = 4 n + 3\\t_{(n+1)}-t_{n} =4\\[/tex]
Step-by-step explanation:
Given:
[tex]t_{n} = 4n - 1[/tex]
To Find:
t₁ = ?
t₂ = ?
t₃ = ?
[tex]t_{(n+1)} = ?\\t_{(n+1)}-t_{n} =?\\[/tex]
Solution:
Put n = 1 in the given equation we get
t₁ = 4×1 - 1 = 3
Put n = 2
t₂ = 4×2 - 1 = 7
Put n = 3
t₃ = 4×3 - 1 = 11
Put n = n + 1
[tex]t_{(n+1)} = 4(n+ 1) -1\\t_{(n+1)} = 4n+4 -1 = 4n+3\\[/tex]
Now we want
[tex]t_{(n+1)}-t_{n} = (4n + 3) - (4n-1)\\t_{(n+1)}-t_{n} = (4n + 3 - 4n+1)\\t_{(n+1)}-t_{n} = 4[/tex]
