A sequence is defined recursively by the given formulas?

aₙ = 4aₙ₋₁ + 1

a₁ = 1 (Given)
When n = 2: a₂ = 4a₁ + 1 = 4×1 + 1 = 5
When n = 3: a₃ = 4a₂ + 1 = 4×5 + 1 = 21
When n = 4: a₄ = 4a₃ + 1 = 4×21 + 1 = 85
When n = 5: a₅ = 4a₄ + 1 = 4×85 + 1 = 341

A recursive sequence definition is really not that difficult once you understand it. I'm going to help you translate the notation into English. To make it clearer, I'm also going to put brackets [] around the subscripted parts.

Recursive definition:
a[1] = 1
a[n] = 4a[n−1] + 1

The first statement basically says, "The first term is 1"
The second statement says, "Every subsequent term is 4 times the prior term, plus 1"

So now just follow the rules:

The first term is 1.

To get to the second term, multiply that by 4 and add 1:
4 * 1 + 1
= 5

To get to the third term, multiply that by 4 and add 1:
4 * 5 + 1
= 21

And repeat the pattern for the fourth term:
4 * 21 + 1
= 85

And the fifth term:
4 * 85 + 1
= 341

Answer:
1, 5, 21, 85, 341

Please try to learn the method rather than "cut and paste".

This gets easier if you understand what the notation means.
a(n) just means the nth term, so, of course a1 means the first term
a(n – 1) means the previous term or (n – 1)th term

Now try to understand this " recursively" part
an = 4*a(n – 1) + 1
It says that the nth term is 4 times the previous term with 1 added on

They gave us the first term. It is a(1) = 1

At this stage you should be thinking what do we need next.
Yes, now we need the second term, but how can we find that ?
Well the idea is that the second term is 4 times the first term + 1
That is quite easy: 4*1 + 1 = 5

By now you know that the second term is called a2, so in notation terms
a2 = 4*a1 + 1 = 4 + 1 = 5
All the stages are like that, so we could write these
a3 = 4*a2 + 1 = 4*5 + 1 = 21
a4 = 4*a3 + 1 = 4*21 + 1 = 85
a5 = 4*a3 + 1 = 4*85 + 1 = 341

That is all there is to it. Now that you can do these, just remember how.

Let's use function notation since it's easier than trying to use subscripts.

I think you have:

a(n) = 4a(n - 1) + 1 where a(1) = 1

So if you are looking for a(4), then we can substitute the value for n = 4 to get:

a(n) = 4a(n - 1) + 1
a(4) = 4a(4 - 1) + 1
a(4) = 4a(3) + 1

But now we need a(3), which we don't yet know, so we go and solve for that:

a(n) = 4a(n - 1) + 1
a(3) = 4a(3 - 1) + 1
a(3) = 4a(2) + 1

We don't know what a(2) is, so we have to solve for that:

a(n) = 4a(n - 1) + 1
a(2) = 4a(2 - 1) + 1
a(2) = 4a(1) + 1

Now that we're here, we know what a(1) is, so we can substitute and solve for a(2). Then work our way backwards to get to a(4), which is what we are looking for:

a(2) = 4(1) + 1
a(2) = 4 + 1
a(2) = 5

Back to the a(3) equation:

a(3) = 4a(2) + 1
a(3) = 4(5) + 1
a(3) = 20 + 1
a(3) = 21

And finally, back to the a(4) equation:

a(4) = 4a(3) + 1
a(4) = 4(21) + 1
a(4) = 84 + 1
a(4) = 85

In computer programming terms, you create a stack of requests until you get to the end condition, which then you pop the requests off of the top of the stack until you've gone through the requests.

a2 = 4a1+1 = 4(1)+1=5
a3=4a2+1 = ...
a4=...
a5=...