Nth Brackets

I did it by hit and trial,

let us see the sum of number in first bracket $(1+2+3+4)=10$

now sum of digits in second bracket $(5+6+7+8)=26$

sum of digits in third bracket $(9+10+11+12)=42$

now put $n=1,2,3$ and see which option gives the above value, I found $(16n-6)$ to be the best option satisfying the above condition

so the correct answer is $\boxed{ 16n-6}$

$•$ this process may be opposed by many people

First digit is $4(n-1) + 1$ . Similary, second, third and fourth digit are $4(n-1) + 2$ , $4(n-1) + 3$ , $4(n-1) + 4$ respectively.

Sum of these terms = $16n - 6$ .

Provided that the question is MCQ, trial and error can also be used.