The correct option is D R>1
R=3065−29653064+2964
∵an−bn=(a−b)(an−1+an−2b+an−3b2+.....+bn−1)
∴R=(30−29)[3064+(3063×29)+....+2964]3064+2964
∵3064+3063×29+...+2964>3064+2964
∴R>1
Hence, option 4.
Alternative approach: Unitary Method
R=(3065−2965)(3064+2964)
If you take this number to be of the form 'n' and 'n+1' instead of 30 and 29, you can solve it using simple numbers like 1(n) and 2(n+1)
i.e (23−13)(22+12)=75>1. Only 1 option satisfies this. Option (d)