Find Pk _{+ 1} for the Given Pk $p _ { k } = \frac { 6 } { k ( k + 1 ) }$

question 183

Multiple Choice

Find pk _{+ 1} for the given pk. $p _ { k } = \frac { 6 } { k ( k + 1 ) }$

• A) \(p _ { k + 1 } = \frac { 2 } { k ( k + 2 ) }\)
• B) \(p _ { k + 1 } = \frac { 6 } { ( k + 1 ) ( k + 1 ) }\)
• C) \(p _ { k + 1 } = \frac { 6 } { ( k + 1 ) ( k + 2 ) }\)
• D) \(p _ { k + 1 } = \frac { 6 } { ( k + 2 ) ( k + 6 ) }\)
• E) \(p _ { k + 1 } = \frac { 6 } { k ( k + 2 ) }\)

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