Answer to Question #306323 in Discrete Mathematics for kasish

Question

Given an=an−1−6an−2 where a0=1 a2=5

a.) list the first 10 terms of the sequence

b.) find a closed form(solve the recurrence relations) 

Answer 1

Solution

Given that

${a_n} = {a_{n - 1}} - 6{a_{n - 2}}\$

And, ${a_0} = 1$ , ${a_2} =5$

For $n=2$

${a_2} = {a_{2 - 1}} - 6{a_{2 - 2}}\\ {a_2} = {a_{1}} - 6{a_{0}}\$

using, ${a_0} = 1$ , ${a_2} =5$

$5 = {a_{1}} - 6(1)\rightarrow {a_{1}}=11$

Similarly, for For $n=3$

${a_3} = {a_{2}} - 6{a_{1}}\\ {a_3} = 5 - 6(11)\\ {a_3} =-61$

Similarly, for For $n=4$

${a_4} = {a_{3}} - 6{a_{2}}\\ {a_4} = -61-6(5)\\ {a_4} =-91$

Continuing in the same way, we get the next few terms, which are tabulated as shown below.