The Solution of the Recurrence Relation in the Previous Problem $c _ { 2 k } = c _ { 0 } / ( 2 k ) , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 1 )$

Multiple Choice

The solution of the recurrence relation in the previous problem is

A) $c _ { 2 k } = c _ { 0 } / ( 2 k ) , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 1 )$
B) $c _ { 2 k } = c _ { 0 } / ( 2 k ) ^ { 2 } , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 1 ) ^ { 2 }$
C) $c _ { 2 k } = c _ { 0 } / ( 2 k ) ! , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 1 ) !$
D) $c _ { 2 k } = c _ { 0 } / ( 2 k + 2 ) ! , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 3 ) !$
E) $c _ { 2 k } = c _ { 0 } / ( 2 k - 1 ) ! , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k ) !$