I wanna find the sum of divisors of a number defined as:$$2^n$$ Let the series be S, $$ so, S = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 +.....2^n $$
Noww, if we factor out the 2 from 2nd term to last term, S can be written like :$$ S= 2^0 + 2(2^0 + 2^1 +2^2 + 2^{(n-1)}) $$
so. $$S = 2^0 + 2S^{(n-1)}$$
I don't know where to go from here :/
Hint: If you let $$S = 2^0+2^1+\cdots+2^{n-1}+2^n$$ then $$2S = 2^1+2^2+\cdots+2^n+2^{n+1}.$$
Subtract the first equation from the second to get $$2S-S = (2^1+2^2+\cdots+2^n+2^{n+1}) - (2^0+2^1+\cdots+2^{n-1}+2^n)$$
You should notice that a lot of terms cancel out. Can you figure out the formula from here?