Series Qn MGMAT Any faster way to do this, and key message?

Hi,
Faster method would be trying to find the pattern.
Sn is 2 times the previous term and then subtracted by some constant.
So, Sn should be having a 2^n since a2^n = a.2.2^(n-1), where 'a' is a constant.
So, once we realize this we can find other things as well.
In this case Sn = a2^n + 2
For n=1, S1 = a.2^(1) + 2 = 3. So, 2a = 1 =>a=1/2
So, Sn = 2^(n-1) + 2
S10 = 2^9 + 2
S9 = 2^8 + 2
So, S10-S9 = 2^9 - 2^8 = 512-256 = 256.

In a similar way, If Sn = 3Sn-1 + constant when Sn will be of the form Sn = a.3^n + b.

My approach was to write out the first few and look for a pattern in the differences, since the question asks for S10 - S9.

The first few are 3, 4, 6, 10, 18.
Differences: 1, 2, 4, 8 ... (powers of 2)

I feel the link below has a wonderful explanation to this problem!

https://www.beatthegmat.com/sequences-t9560.html#38369