Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.

Step 1:

In this problem we have to find the solution for these recurrence relation, where we given the initial condition.

a: the recurrence relation is given as

an = 3 an-1 with a0= 2

Now we starting with initial condition a0 = 2 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= 3a1-1

a1 = 3 a0 = 3.2

a2 = 3a2-1

a2 = 3a1 = 3.3.2 = 32.2

a3 = 3a3-1

a3 = 3a2 = 3.32.2 = 33.2

………..

………..

an = 3an-1 = 3n.2

Step 2:

b: the recurrence relation is given as

an = an-1 +2 with a0= 3

Now we starting with initial condition a0 = 3 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+2

a1 = a0 +2 = 3+2

a2 = a2-1 +2

a2 = a1 +2 = 3+2+2 = 3+2.2

a3 = a3-1+2

a3 = a2 +2 = 3+2.2+2 = 3+3.2

………..

………..

an = an-1 +2 = 3+ 2n.

Step 3:

c: the recurrence relation is given as

an = an-1 +n with a0= 1

Now we starting with initial condition a0 = 1 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+1

a1 = a0 +1 = 1+1

a2 = a2-1 +2

a2 = a1 +2 = 1+1+2 = 1+ (1+2)

a3 = a3-1+3

a3 = a2 +3 = 1+1+2+3 = 1+ (1+2+3)

a4 = a3 +4 = 1+1+2+3+4 = 1 + (1+2+3+4 ) = 1+ n(n+1)/2

(since by A.P. series 1+2+3+.........+n = n(n+1)/2)

………..

………..

an = an-1 + n = 1+ n(n+1)/2.

Step 4:

d: the recurrence relation is given as

an = an-1 + 2n +3 with a0= 4

Now we starting with initial condition a0 = 4 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+2.1 +3

a1 = a0 +2.1+3 = 4+ 2.1 +3

a2 = a2-1 +2.2 +3

a2 = a1 +2.2+3 = 4 +2.1+3+2.2+3 = 4 + 2(1+2) +2.3

a3 = a3-1+2.3+3

a3 = a2 +2.3+3 =4 +2 (1+2+3) +3.3

………..

………..

an = an-1 + 2n +3 = 4 +2( 1+2+...... +n) +n.3= 4+ 2n( n+1)/2 +n.3,

(since by A.P. series 1+2+3+.........+n = n(n+1)/2)

an = an-1 + 2n +3 = n2 +4n +4.