Find the first five terms of these recursively-defined sequences:

Question

1) a₁ = 2,   a_n = 2a_n-1 – 5

2) a₁ = 3, a_n = _an-1  + 3     

3) a₁ = 2, a₂ = 4, a_n = a_n-1 + a_n-2

Answer 1

1) a_1=2, a_n=2a_{n-1}-5

When n=2, using the equation, we have a_2=2a_{2-1}-5

So a_2=2a_1-5

We know a_1=2 because that was given. So a_2=2*2-5=4-5=-1

So a_2=-1

Now when n=3, we have a_3=2a_{3-1}-5.

So a_3=2a_2-5

Now plug in a_2 to get a_3=2(-1)-5=-2-5=-7

So a_3=-7

When n=4, we have a_4=2a_{4-1}-5.

So a_4=2a_3-5

Now plug in a_3 to get a_4=2(-7)-5=-14-5=-19

So a_4=-19

When n=5, we have a_5=2a_{5-1}-5.

So a_5=2a_4-5

Now plug in a_4 to get a_5=2(-19)-5=-38-5=-43

So a_5=-43

2) a_1=3, a_n=a_{n-1}+3

When n=2, using the equation, we have a_2=a_{2-1}+3

So a_2=a_1+3

We know a_1=3 because that was given. So a_2=3+3=6

So a_2=6

Now when n=3, we have a_3=a_{3-1}+3

So a_3=a_2+3

Now plug in a_2 to get a_3=6+3=9

So a_3=9

When n=4, we have a_4=a_{4-1}+3

So a_4=a_3+3

Now plug in a_3 to get a_4=9+3=12

So a_4=12

When n=5, we have a_5=a_{5-1}+3

So a_5=a_4+3

Now plug in a_4 to get a_5=12+3=15

So a_5=15

3) a_1=2, a_2=4, a_n=a_{n-1}+a_{n-2}

When n=3, using the equation, we have a_3=a_{3-1}+a_{3-2}

So a_3=a_2+a_1

Now plug in a_1=2 and a_2=4 (which were given) to get a_3=4+2=6

So a_3=6

Now when n=4, we have a_4=a_{4-1}+a_{4-2}

So a_4=a_3+a_2

Now plug in a_2=4 and a_3=6 to get a_4=6+4=10

So a_4=10

When n=5, we have a_5=a_{5-1}+a_{5-2}

So a_5=a_4+a_3

Now plug in a_3=6 and a_4=10 to get a_5=10+6=16

So a_5=16