Answer:
a) a₁ = 14, a₂ = 13, a₃ = 12, a₄ = 11
b) an+4 = 16
c) Does not exist
Step-by-step explanation:
The last digit in student number is given as 7.
a) a₀ = 7
Since an < 10 we use 2an
Therefore a₁ = a₀₊₁ = 2 × a₀ = 2 × 7 = 14
a₂ = a₁₊₁ = a₁ - 1 = 14 -1 = 13
a₃ = a₂₊₁ = a₂ - 1 = 13 -1 = 12
a₄ = a₃₊₁ = a₃ - 1 = 12 -1 = 11
b) an = 1, we have an+1 = 2an
Therefore an+2 = an+1+1 = 2 × 2 = 4
an+3 = an+2+1 = 2 × 4 = 8
an+4 = an+3+1 = 2 × 8 = 16
Therefore, an+4 = 16
c) If a₀ = 3, therefore a₁ = a₀₊₁ = 2×3 = 6
a₂ = a₁₊₁ = 2×6 = 12
a₃ = a₂₊₁ = a₂₋₁ = 12- 1 =11
a₄ = a₃₊₁ = a₃₋₁ = 11- 1 =10
a₅ = a₄₊₁ = a₄₋₁ = 10- 1 = 9
a₆ = 2×9 = 18
We can therefore see that limn→[infinity] does not exist.
