Given that \(A_{n} = 2A_{n−1} + 3\) for all \(n ≥ 1\) and \(A_{4} = 45\) and we need to find the value of \(A_{1}\))

Let \(A_{n} = 2A_{n−1} + 3\) be denoted by (1)
So, we need to use the above equation and try to calculate \(A_{3}\) from \(A_{4}\) and then \(A_{2}\) from \(A_{3}\) and eventually \(A_{1}\) from \(A_{2}\).
Two ways of doing this

Method 1: Find value at each stage and then solve for the next stage

So, lets start with the labor work!
Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\)
=> \(45 = 2A_{3} + 3\)
=> \(2A_{3} = 42\) => \(A_{3} = 21\)
Putting n = 3 in (1)we get, \(A_{3} = 2A_{3−1} + 3\)
=> \(21 = 2A_{2} + 3\)
=> \(2A_{2} = 18\) => \(A_{2} = 9\)
Putting n = 2 in (1) we get, \(A_{2} = 2A_{2−1} + 3\)
=> \(9 = 2A_{1} + 3\)
=> \(2A_{1} = 6\) => \(A_{1} = 3\)

So, Answer will be E.

Method 2: Simplify the entire expression first and then solve for the value

Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\) => \(A_{4} = 2A_{3} + 3\), Now let's find \(A_{3}\)
Putting n=3 in (1) we get, \(A_{3} = 2A_{3−1} + 3\) => \(A_{3} = 2A_{2} + 3\)
Similarly, \(A_{2} = 2A_{1} + 3\)
=> \(A_{4} = 2A_{3} + 3\) will give us \(A_{4} = 2(2A_{2} + 3) + 3\) => \(A_{4} = 4A_{2} + 9\)
And this will give us \(A_{4} = 4(2A_{1} + 3) + 9\) => \(A_{4} = 8A_{1} + 21\)
=> \(45 = 8A_{1} + 21\)
=> \(8A_{1} = 24\)
=> \(A_{1} = 3\)

So, Answer will be E.
Hope it helps!

Watch the following video to learn How to Sequence problems