Find Pk+1 for the given Pk. $P _ { k } = \frac { 6 } { k ( k + 1 ) }$
A) $P _ { k + 1 } = \frac { 6 } { k ( k + 1 ) } + 1$
B) $P _ { k + 1 } = \frac { 6 } { k ( k + 1 ) } + \frac { 6 } { ( k + 1 ) ( k + 2 ) }$
C) $P _ { k + 1 } = \frac { 6 } { ( k + 1 ) ( k + 2 ) }$
D) $P _ { k + 1 } = \frac { 6 } { k ( k + 2 ) }$
E) $P _ { k + 1 } = \frac { 36 } { ( k + 1 ) ( k + 2 ) }$

Question

Quizplus · Accepted Answer

To find $P_{k+1}$, we substitute $k+1$ for $k$ in the given formula $P_k = \frac{6}{k(k+1)}$, resulting in $P_{k+1} = \frac{6}{(k+1)((k+1)+1)} = \frac{6}{(k+1)(k+2)}$.

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