0 Đ( ) 22( )= − ≤ ≤KN N C −− K N1 2N( 1 ) ( N0 2 N1 2 N3 3

2,0 đ

( )

2

2

( )

= − ≤ ≤

k

n n C

k n

1 2

n

( 1 ) (

n

0

2

n

1

2

n

3

3

...

n

n

2

2

)

S = n nC

+ C

+ C

+ + C

0,5 đ

( 1 .2 )

n

2

S = n n

0,5 đ

b)