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

question 15

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

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