- #1
taya
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Solve each of the following linear recurrence relations:
1. t(1)=1 t(2)=4
t(n) - 5t(n-1) + 6t (n-2)= 0 for n>1
2. a(n)=4a(n-1) - 4a (n-2)
with initial conditions a(0) = 4 and a(1)=12
3. t(1)=3 t(2)=3
t(n) + 2t (n-1) + t(n-2) = 0
1. t(1)=1 t(2)=4
t(n) - 5t(n-1) + 6t (n-2)= 0 for n>1
2. a(n)=4a(n-1) - 4a (n-2)
with initial conditions a(0) = 4 and a(1)=12
3. t(1)=3 t(2)=3
t(n) + 2t (n-1) + t(n-2) = 0