When a ♁ B = n (n is a constant), we can get (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n + 2, then (a + 2) ♁ (B + 2)= (can you talk about the process? Thank you. Thank you very much. The sooner the better.)

When a ♁ B = n (n is a constant), we can get (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n + 2, then (a + 2) ♁ (B + 2)= (can you talk about the process? Thank you. Thank you very much. The sooner the better.)

a♁b=n
We obtain (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n + 2,
It can be seen that (a + 1) ♁ (B + 1) = n + 1 + 2 = n + 3
The results increase with each increase of a and B
So (a + 2) ♁ (B + 2) = n + 1 * 2 + 2 * 2 = n + 6