Question #306323

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) 


Expert's answer

Solution


Given that


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


And, a0=1{a_0} = 1 , a2=5{a_2} =5


For n=2n=2


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


using, a0=1{a_0} = 1 , a2=5{a_2} =5


5=a1−6(1)→a1=115 = {a_{1}} - 6(1)\rightarrow {a_{1}}=11


Similarly, for For n=3n=3

a3=a2−6a1a3=5−6(11)a3=−61{a_3} = {a_{2}} - 6{a_{1}}\\ {a_3} = 5 - 6(11)\\ {a_3} =-61


Similarly, for For n=4n=4

a4=a3−6a2a4=−61−6(5)a4=−91{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.









LATEST TUTORIALS
APPROVED BY CLIENTS