help plz

(3√3 +5)²ⁿ⁺¹ =α+β

let (3√3-5)²ⁿ⁺¹ =f

now f(α+β)= (3√3-5)²ⁿ⁺¹ X (3√3 +5)²ⁿ⁺¹ = 2²ⁿ⁺¹

=> f = [2/ (3√3 +5)]²ⁿ⁺¹ <1 [as manish bhaiya told]

now α+β+f = (3√3 +5)²ⁿ⁺¹ + (3√3-5)²ⁿ⁺¹

=> α+β+f = 2 (²ⁿ⁺¹C₀ (3√3)²ⁿ⁺¹ + .............+ 5²ⁿ⁺¹) = not an integer

=> but, α+β+f = I (intger) + F(fraction)

this I will obviously be an even integer (2 is there in the front :P)

and I is divisible by 10.

but i dunno how the prev integer is divisible by 3 ....

in fact, it is the next integer that is divisible by 3...

answer is 3