Given:
The integer n≥1.
Calculation:
Consider; 1−12=12.
Then,
(1− 1 2)n=( 1 2)n( n 0 )(1)n(− 1 2)0+( n 1 )(1)n−1(− 1 2)1+( n 2 )(1)n−2(− 1 2)2+( n 3 )(1)n−3(− 1 2)3+…+( n n−1 )(1)1(− 1 2)n−1+( n n )(1)0(− 1 2)n=( 1 2)n( n 0 )−12( n 1 )+122( n 2 )−123( n 3 )+…+(−1)n−112 n−1( n n−1 )=( 1 2)n−(−1)n( 1 2)n←(1)
Let n=2m where m∈ℤ+∪0 ,
From (1) ;
Right side=(12)2m−(−1)2m(12)2m=(12)2m−(12)2m=0←(2)
Hence, from 1 and 2;
(n0)−12(n1)+122(n2)−123(n3)+…+(−1)n−112n−1(n n−1)=(12)n−(−1)n(12)n=0 when n is an even integer