163 Discrete Mathematics Review 2 Use the following to answer ...

10. 1 2 ,2 2 ,3 3 ,4 2 ,….
Solution. We see that 2 , 1,2,
2 2
an
= n n= … Therefore a − n 1
a = n
( n + 1) − +
n = 2n
+ 1
and the sequence can be described recursively as
a = 1, a = a + 2n+
1.
1 n+
1
n
Use the following to answer question 11:
In the questions below solve the recurrence relation either by using the characteristic equation
or by discovering a pattern formed by the terms.
11. a n = −10a n − 1 − 21a n − 2 , a 0 = 2, a 1 = 1.
Solution. The characteristic equation of the above recurrence relation is
2
r + 10r+ 21 = 0 .
The left part can be factored as ( r+ 3)( r+ 7) whence the solutions of the characteristic
equation are r1 =− 3, r2
=−7
, and the general solution of the recurrence relation is
n
n
an
= α1( − 3) + α2( − 7) .
From the initial conditions we find
α1+ α2
= 2
3α1+ 7α2
=− 1
.
By multiplying the first equation by 7 and subtracting from it the second equation we
get 4α 1
= 15whence α = 15
1
4
and α = 2− α = 2− 15 =− 7 . Finally
2 1
4 4
15 n 7
a
n
= ( −3) − ( − 7)
4 4
n
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