If n is a positive integer and `(3sqrt(3)+5)^(2n+1)=l+f` where l is an integer annd `0 lt f lt 1`, then

Options

A. `alpha ` is an even integer
B. `(alpha + beta)^(2)` is divisible by `2^(2n+1)`
C. the integer just below `(3sqrt(3) + 5)^(2n+1)` divisible by 3
D. `alpha ` is divisible by 10

Answer

a,d

Explanation

`(3sqrt(3) + 5)^(2n +1) = (sqrt(27) + 5)^(2n +1)` …(i)
Now , let ` alpha + beta = (sqrt(27) + 5)^(2n+1)` …(ii)
` 0 lt beta ' lt 1` …(iii)
and let ` beta' = (sqrt(27) - 5)^(2n+1)` …(iv)
` 0 lt beta' lt 1 ` …(iv)
On subtracting Eq. (iii) from Eq. (i) , we get
` alpha + beta - beta = (sqrt(27) + 5)^(2n+1) - (sqrt(27)- 5)^(2n +1)` ...(v)
`rArr alpha + 0 = 2p ` (even integer) , ` AA p in ` N
` rArr alpha = 2p ` = even integer
Also , from Eq . (v) , we grt
` alpha = (sqrt(27) + 5)^(2n+1) - (sqrt(27) -5)^(2n+1)` divisible by
` (sqrt(27) +5) - (sqrt(27) - 5) ` . i.e. divisible by 10.