Problem: ∑ k = 0 49 ( − 1 ) k ( 99 2 k ) ∑ k = 0 4 9 ( − 1 ) k ( 2 k 9 9 ) ( n j ) = n ! j ! ( n − j ) ! ( j n ) = j ! ( n − j ) ! n !
Answer Choices:
A. − 2 50 − 2 5 0
B. − 2 49 − 2 4 9
C. 0 0
D. 2 49 2 4 9
E. 2 50 2 5 0
Solution:
By the Binomial Theorem,
( 1 + i ) 99 = ( 99 0 ) + ( 99 1 ) i + ( 99 2 ) i 2 + ( 99 3 ) i 3 + ⋯ + ( 99 99 ) i 99 ( 1 + i ) 9 9 = ( 0 9 9 ) + ( 1 9 9 ) i + ( 2 9 9 ) i 2 + ( 3 9 9 ) i 3 + ⋯ + ( 9 9 9 9 ) i 9 9
Note that the real part of this series is
( 99 0 ) − ( 99 2 ) + ( 99 4 ) − ( 99 6 ) + ⋯ − ( 99 98 ) ( 0 9 9 ) − ( 2 9 9 ) + ( 4 9 9 ) − ( 6 9 9 ) + ⋯ − ( 9 8 9 9 )
the very sum we want to find. Since ( 1 + i ) = 2 ( cos π 4 + i sin π 4 ) ( 1 + i ) = 2 ( cos 4 π + i sin 4 π ) ( 1 + i ) 99 = 2 ( 99 / 2 ) ( cos 99 π 4 + i sin 99 π 4 ) ( 1 + i ) 9 9 = 2 ( 9 9 / 2 ) ( cos 4 9 9 π + i sin 4 9 9 π ) ( 1 + i ) 99 ( 1 + i ) 9 9 2 99 / 2 cos 99 π 4 = 2 99 / 2 cos 3 π 4 = 2 99 / 2 ( − 2 2 ) = − 2 49 2 9 9 / 2 cos 4 9 9 π = 2 9 9 / 2 cos 4 3 π = 2 9 9 / 2 ( − 2 2 ) = − 2 4 9