help plz

is it multiple choice??

i am getting 3 answers... 1,3,4.

no no no no sryy..........

mistake!!

(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...

yeah.... dats true...

but bhaiya :P

ans kuchh aur hai.... ans nahi mila....

well... then,

a+b - f = 2m (an even integer)

so b-f = 2m - a = intgr

0<=b<1 , 0<=f<1

so b-f =0 => a = 2m

but then again it is an even integer...

and the last term, will not have any 3 in it... so how can it be a multiple of 3 //

help :'(