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Solve the following recurrence relation : ar−3ar−1=2,r≥1,a0=1
1 Answer
written 2.9 years ago by |
Solution:
The characteristic equation is,
(λ−3)=0λ=3
Hence the homogeneous solution is,
a(h)r=A(3)r
Particular solution is of the type P (constant),
ar=P
ar−1=P
Substituting the value of ar and ar−1 in the given recurrence relation.
P−3P=2−2P=2P=−1ar=A(3)r−1
Using initial condition,
a0=1
1=A−1⇒A=2ar=2(3r)−1