- #1
erisedk
- 374
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Homework Statement
The integer next to (√3 + 1 )^2n is -- (n is a natural number)
Ans: Divisible by 2^(n+1)
Homework Equations
The Attempt at a Solution
(√3 + 1 )^2n will have an integral and a fractional part.
So, I + f = (√3 + 1 )^2n
(√3 - 1 )^2n will always be fractional as (√3 - 1) < 1
So, f' = (√3 - 1 )^2n
On adding I + f and f', all the irrational terms (which are also the even terms) will cancel out and the odd terms will get added.
I + f + f' = 2 [ (2nC0 √3^2n) + (2nC2 √3^2n-2) + ... ]
I + f + f' = 2 [ (2nC0 3^n) + (2nC2 3^n-1) + ... ]
Since I + f + f' is an even integer, 0<f<1, 0<f'<1 and 0<f+f'<2
f+f'=1
So, I = 2 [ (2nC0 3^n) + (2nC2 3^n-1) + ... ] - 1
I need the integer next to (√3 + 1 )^2n. Let the required integer be denoted by I'.
I' = 2[ (2nC0 3^n) + (2nC2 3^n-1) + ... ] - 1 + 1
I' =2 [ (2nC0 3^n) + (2nC2 3^n-1) + ... ]
From here, the only thing I infer is that I' is divisible by 2. How do I show that this is divisible by 2^(n+1) ?