Find pk_{+ 1} for the given p_k. $p _ { k } = \frac { k } { 5 } ( 8 k + 1 )$ A) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 5 )$ B) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 6 )$ C) $p _ { k + 1 } = \frac { k + 1 } { 6 } ( 8 k + 6 )$ D) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 9 )$ E) $p _ { k + 1 } = \frac { k + 1 } { 6 } ( 8 k + 9 )$

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The Answer of Find pk_{+ 1} for the given p_k. $p _ { k } = \frac { k } { 5 } ( 8 k + 1 )$ A) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 5 )$ B) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 6 )$ C) $p _ { k + 1 } = \frac { k + 1 } { 6 } ( 8 k + 6 )$ D) $p _ { k + 1 } = \frac { k + 1 } { 5 } ( 8 k + 9 )$ E) $p _ { k + 1 } = \frac { k + 1 } { 6 } ( 8 k + 9 )$

Find Pk + 1 for the Given Pk pk=k5(8k+1)p _ { k } = \frac { k } { 5 } ( 8 k + 1 )pk​=5k​(8k+1)

Find Pk_{+ 1} for the Given P_k $p _ { k } = \frac { k } { 5 } ( 8 k + 1 )$