Find pk_{+ 1} for the given pk. $p _ { k } = \frac { 1 } { 4 ( k + 2 ) }$ A) $p _ { k + 1 } = \frac { 1 } { k ( k + 3 ) }$ B) $p _ { k + 1 } = \frac { 4 } { ( k + 1 ) ( k + 2 ) }$ C) $p _ { k + 1 } = \frac { 4 } { k ( k + 3 ) }$ D) $p _ { k + 1 } = \frac { 1 } { ( k + 3 ) ( k + 2 ) }$ E) $p _ { k + 1 } = \frac { 1 } { 4 ( k + 3 ) }$

Question

Quizplus · Accepted Answer

The given formula for pk is $p _ { k } = \frac { 1 } { 4 ( k + 2 ) }$. To find $p _ { k + 1 }$, replace k with k+1: $p _ { k + 1 } = \frac { 1 } { 4 ( (k+1) + 2 ) } = \frac { 1 } { 4 ( k + 3 ) }$.

Find Pk + 1 for the Given Pk pk=14(k+2)p _ { k } = \frac { 1 } { 4 ( k + 2 ) }pk​=4(k+2)1​

Find Pk_{+ 1} for the Given Pk $p _ { k } = \frac { 1 } { 4 ( k + 2 ) }$