(1, 2)4000 = (1 + 0.2)4000
Now, expanding by binomial theorem, we get
(1 + 0.2)4000 = 4000 C0 (1)4000 (0.2)0 + 4000 C1 (14000 – 1 (0.2)1 +4000 C3 (1)4000 – 3 (0.2)3 + …
= 1 + 4000(0.2) + other terms of the expansion
= 1 + 800 + other terms of the expansion
= 801 + other terms of the expansion
Hence, (1, 2)4000 > 800