Question

Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions.

a) an=6an-1, a0=2
b) an=a2n-1, a1=2
c) an=an-1+3an-2, a0=1 , a1=2
d) an=nan-1+n2 an-2 a0=1 ,a1=1
e) an= an-1+an-3 , a0=1, a1=2 , a2=0

Answer 1

See explanation

Explanation:

Solving (a):

`a_n = 6a_(n-1)` where
`a_0 = 2`

n = 1

`a_n = 6a_(n-1)`

`a_1 = 6a_(1-1)`

`a_1 = 6a_(0)`

Substitute 2 for
`a_0`

`a_1= 6 * 2`

`a_1= 12`

n = 2

`a_n = 6a_(n-1)`

`a_2 = 6a_(2-1)`

`a_2 = 6a_(1)`

Substitute 12 for
`a_1`

`a_2= 6 * 12`

`a_2= 72`

n = 3

`a_n = 6a_(n-1)`

`a_3 = 6a_(3-1)`

`a_3 = 6a_2`

Substitute 72 for
`a_2`

`a_3= 6 * 72`

`a_3= 432`

n = 4

`a_n = 6a_(n-1)`

`a_4 = 6a_(4-1)`

`a_4 = 6a_(3)`

Substitute 432 for
`a_3`

`a_4 = 6 * 432`

`a_4 = 2592`

n = 5

`a_n = 6a_(n-1)`

`a_5 = 6a_(5-1)`

`a_5 = 6a_4`

Substitute 2592 for
`a_4`

`a_5 = 6 * 2592`

`a_5 = 15552`

n = 6

`a_n = 6a_(n-1)`

`a_6 = 6a_(6-1)`

`a_6 = 6a_(5)`

Substitute 15552 for
`a_5`

`a_6 = 6 * 15552`

`a_6 = 93312`

`a_1= 12`
`a_2= 72`
`a_3= 432`
`a_4 = 2592`
`a_5 = 15552`
`a_6 = 93312`

Solving (b):

`a_n = a^2_{n - 1` where
`a_1 = 2`

We have

`a_1 = 2` which serves as the first term

n =2

`a_n = a^2_{n - 1`

`a_2 = a^2_(2-1)`

`a_2 = a^2_(1)`

Substitute 2 for
`a_1`

`a_2 = 2^2`

`a_2 = 4`

n = 3

`a_3 = a^2_(3-1)`

`a_3 = a^2_(2)`

`a_3 = 4^2`

`a_3 = 16`

n = 4

`a_4 = a^2_(4-1)`

`a_4 = a^2_3`

`a_4 = 16^2`

`a_4 = 256`

n =5

`a_5 = a^2_{5-1`

`a_5 = a^2_4`

`a_5 = 256^2`

`a_5 = 65536`

n = 6

`a_6 = a^2_{6-1`

`a_6 = a^2_{5`

`a_6 = 65536^2`

`a_6 = 4294967296`

`a_1 = 2`
`a_2 = 4`
`a_3 = 16`
`a_4 = 256`
`a_5 = 65536`
`a_6 = 4294967296`

Solving (c):

`a_n=a_(n-1)+3a_(n-2);`
`a_0=1` ; ;
`a_1=2`

`a_1=2` ---- First term

n = 2

`a_n=a_(n-1)+3a_(n-2);` becomes

`a_2=a_(2-1)+3a_(2-2)`

`a_2=a_1+3a_0`

Substitute values for a1 and a0

`a_2=2+3 * 1`

`a_2=2+3`

`a_2=5`

n = 3

`a_n=a_(n-1)+3a_(n-2);` becomes

`a_3=a_(3-1)+3a_(3-2)`

`a_3=a_(2)+3a_(1)`

`a_3=5+3 * 2`

`a_3=5+6`

`a_3=11`

n = 4

`a_n=a_(n-1)+3a_(n-2);` becomes

`a_4=a_(4-1)+3a_(4-2)`

`a_4=a_(3)+3a_(2)`

`a_4=11+3 * 5`

`a_4=11+15`

`a_4=26`

n = 5

`a_n=a_(n-1)+3a_(n-2);` becomes

`a_5=a_(5-1)+3a_(5-2);`

`a_5=a_(4)+3a_3`

`a_5=26+3 * 11`

`a_5=26+33`

`a_5=59`

n = 6

`a_n=a_(n-1)+3a_(n-2);` becomes

`a_6=a_(6-1)+3a_(6-2)`

`a_6=a_(5)+3a_4`

`a_6=59+3*26`

`a_6=59+78`

`a_6=137`

`a_1=2`
`a_2=5`
`a_3=11`
`a_4=26`
`a_5=59`
`a_6=137`

Solving (d):

`a_n=na_(n-1)+n^2a_(n-2)`;
`a_0=1`;
`a_1=1`

`a_1=1` --- First term

n = 2

`a_n=na_(n-1)+n^2a_(n-2)` becomes

`a_2=2 * a_(2-1)+2^2a_(2-2)`

`a_2=2 * a_1+4*a_0`

`a_2=2 * 1+4*1`

`a_2=2 +4`

`a_2=6`

n = 3

`a_n=na_(n-1)+n^2a_(n-2)` becomes

`a_3=3 * a_(3-1)+3^2 * a_(3-2)`

`a_3=3 * a_(2)+9 * a_(1)`

`a_3=3 * 6+9 * 1`

`a_3=18+9`

`a_3=27`

n = 4

`a_n=na_(n-1)+n^2a_(n-2)` becomes

`a_4=4*a_(4-1)+4^2*a_(4-2)`

`a_4=4*a_(3)+16*a_(2)`

`a_4=4*27+16*6`

`a_4=204`

n = 5

`a_n=na_(n-1)+n^2a_(n-2)` becomes

`a_5=5 * a_(5-1)+5^2 * a_(5-2)`

`a_5=5 * a_(4)+25 * a_(3)`

`a_5=5 * 204+25 *27`

`a_5=1695`

n = 6

`a_n=na_(n-1)+n^2a_(n-2)` becomes

`a_6=6 * a_(6-1)+6^2*a_(6-2)`

`a_6=6 * a_(5)+36*a_(4)`

`a_6=6 * 1695+36*204`

`a_6=17514`

`a_1=1`
`a_2=6`
`a_3=27`
`a_4=204`
`a_5=1695`
`a_6=17514`

Solving (e):

`a_n= a_(n-1)+a_(n-3);\ a_0=1; a_1=2; a_2=0`

First term:
`a_1=2`

Second Term:
`a_2=0`

n = 3

`a_n= a_(n-1)+a_(n-3)` becomes

`a_3= a_(3-1)+a_(3-3)`

`a_3= a_(2)+a_0`

`a_3= 0+1`

`a_3= 1`

n = 4

`a_n= a_(n-1)+a_(n-3)` becomes

`a_4= a_(4-1)+a_(4-3)`

`a_4= a_(3)+a_(1)`

`a_4= 1+2`

`a_4=3`

n = 5

`a_n= a_(n-1)+a_(n-3)` becomes

`a_5= a_(5-1)+a_(5-3)`

`a_5= a_(4)+a_(2)`

`a_5= 3 + 0`

`a_5= 3`

n = 6

`a_n= a_(n-1)+a_(n-3)` becomes

`a_6= a_(6-1)+a_(6-3)`

`a_6= a_(5)+a_(3)`

`a_6= 3 + 1`

`a_6= 4`

`a_1=2`
`a_2=0`
`a_3= 1`
`a_4=3`
`a_5= 3`
`a_6= 4`