Binomial Theorem Part 1

General term =‌256Cr⋅(√3)256−r⋅(‌8√5)r
=‌256Cr⋅(3)‌

256−r

2

⋅(5)‌

5

8

When ‌

256−r

2

is integer then (3)‌

256−r

2

is integer.
And when ‌

r

8

is integer then (5)‌

r

8

is integer.
Entire general term will be integer when ‌

256−r

2

and ‌

r

8

both are integer.
‌

256−r

2

is integer when r=0,2,4,6,.....,256
‌

r

8

is integer when r=0,8,16 256
Now both ‌

256−r

2

and ‌

r

8

will be integer when r=0,8,16,....,256 (This is an AP)
∴256=0+(n−1)8 using formula of AP,tn=a+(n−1)d
∴n=‌

256

8

+1=32+1=33