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) $c _ { 2 k } = c _ { 0 } / ( 2 k - 1 ) ! , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k ) !$

Question

Quizplus · Accepted Answer

The correct solution involves factorials in the denominator, which is typical for series solutions of differential equations or generating functions in combinatorics, where the factorial accounts for the number of ways to arrange elements.

The Solution of the Recurrence Relation in the Previous Problem c2k=c0/(2k),c2k+1=c1/(2k+1)c _ { 2 k } = c _ { 0 } / ( 2 k ) , c _ { 2 k + 1 } = c _ { 1 } / ( 2 k + 1 )c2k​=c0​/(2k),c2k+1​=c1​/(2k+1)

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