Computation complexity with simple algebra expression reduction

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I'm watching this computer science video on computational time complexity of a function where they introduce some maths and it doesn't make sense to me. I'm not even sure what the name for this maths is so I apologise for that.

Basically they start with the the function T(n) as defined on line 1 and on line 2 it's given that t(0) = 1:

1    T(n) = T(n-1) + 3 if n > 0
2    T(0) = 1

Then they say that the above expression can be reduced as:

3    T(n) = T(n-1) + 3
4         = T(n-2) + 6
5         = T(n-3) + 9
6         = T(n-k) + 3k

The "reduction" part is which I don't understand - lines 3 to 5. They say "t(n-1) can be written as T(n-2) + 3 so overall the expression will be T(n-2) + 6". That is what I don't get.

So, from the above, why can line 3 be written as it is on line 4? And then again why can line 4 can be written as it is on line 5. They seem to be applying some trivial rules that I can't remember from high school.

Any explanation as to what they are doing is welcome. Also, links to where I can learn/practice more of this type of problem would be great.

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Essentially they just keep applying the first rule over and over.

Since $T(n)=T(n-1)+3$ for $n>0$ then if you want to rewrite $T(n-1)$ you just think of $n-1$ as being your $n$. It might be easier to see if we work with $T(k-1)$ then if you call $k-1=n$ then you have $T(k-1)=T(n)=T(n-1)+3=T((k-1)-1)+3$ where we just plug in for $n=k-1$. The same approach using iterations on $n$ gives $T(n-1)=T((n-1)-1)+3=T(n-2)+3=(T((n-2)-1)+3)+3=T(n-3)+6$.

The essential bit is that we say $T(n)=T(n-1)+3$ for all $n>0$. It might be even easier to see with some actual numbers.

If you have $T(5)=T(5-1)+3=T(4)+3=(T(4-1)+3)+3=T(3)+6$.

Does that explain it?

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On

It can be thought of as follows

$T(n)=T(n-1)+3$---------exp 1.

Now substituting n-1 in place of n in exp 1

$T(n-1)=T(n-2)+3$--------exp 2

Therefore,

$T(n)=T(n-1)+3=(T(n-2)+3)+3=T(n-2)+6$ and so on....$T(n)$ can be written as $T(n)=T(n-k)+3k$