The greatest integer less than or equal to `(sqrt2+1)^6` is

Options

A. 197
B. 198
C. 196
D. 199

Explanation

Let `(2sqrt(2) +1)^(6) = I + F` where I `in N and 0 lt F lt 1`.
Clearly , I is the greatest integer less then or equal to `(2 sqrt(2) +1)^(6)`
Let G =` (sqrt(2) -1)^(6)` . Then ,
`I + F + G = (sqrt(2) + 1)^(6) + (sqrt(2) -1)^(6)`
`rArr I + F + G = 2 {""^(6)C_(0) (sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) +....}`
`rArr I + F + G ` = An even integer
`rArr F + G = 1`
Again, `I + F + G`
`2={""^(6)C_(0)(sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) + ""^(6)C_(4) (sqrt(2))^(6-4) + ""^(6)C_(6) (sqrt(2))^(6-6)}`
`rArr I + 1 = 2 (8 + 15 xx4 + 15 xx4 + 15 xx2 +1) [because F + G = 1 ]`
`rArr I = 197`