@shesh_neo : You have to form a recurrence relation and solve it and apply initial conditions to get to the closed form . Let me know if you have trouble forming a recurrence relation , I will post the relation .

for the part 1:
let’s say we have f(n) for n, so we can easily deduce, f(n+1)=f(n)*3+2^n-1
so, f(n+1)+2^(n+1)-1/2=(f(n)+2^n-1/2)*3
let’s say k(n)=f(n)+2^n-1/2
so k(1)=3/2, k(n+1)=k(n)*3, so k(n)=3^n/2
so, f(n)=3^n/2-2^n+1/2

@sparshgunner12 : check my solution for the recursive relation and the explanation about how to come up with the formula. Everything is mentioned in the comments.

@shesh_neo : check my solution for the recursive relation and the explanation about how to come up with the formula. Everything is mentioned in the comments.

To all the people trying this question, here’s a special note:
Don’t calculate like, a=4^n%MOD, b= 2^n%MOD and c=3^n%MOD, and then do 2*a + b/2 + b + ((c+1)/2), and similar things for R1… This will give you a wrong answer…because what we actually want to find is like : ((3^n + 1)/2)%MOD, which is Actually : (3^n + 1)%MOD * (2^MOD-2)%MOD… So don’t forget to leave the divider alone, take that into consideration too…I did this mistake many a times before realizing it !
Cheers !