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) $p _ { k + 1 } = \frac { 6 } { k ( k + 2 ) }$

Question

Quizplus · Accepted Answer

To find $p_{k+1}$, we substitute $k+1$ for $k$ in the given formula for $p_k$, which is $\frac{6}{k(k+1)}$. So, $p_{k+1} = \frac{6}{(k+1)((k+1)+1)} = \frac{6}{(k+1)(k+2)}$, which matches choice C.

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

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