Find Pk₊₁ for the given P_k. $P _ { k } = \frac { 2 } { k ( k + 1 ) }$ A) $P _ { k + 1 } = \frac { 2 } { k ( k + 2 ) }$ B) $P _ { k + 1 } = \frac { 2 } { k ( k + 1 ) } + \frac { 2 } { ( k + 1 ) ( k + 2 ) }$ C) $P _ { k + 1 } = \frac { 4 } { ( k + 1 ) ( k + 2 ) }$ D) $P _ { k + 1 } = \frac { 2 } { ( k + 1 ) ( k + 2 ) }$ E) $P _ { k + 1 } = \frac { 2 } { k ( k + 1 ) } + 1$

Question

Quizplus · Accepted Answer

The expression for $P_{k+1}$ is derived by replacing $k$ with $k+1$ in the original formula, resulting in $\frac{2}{(k+1)((k+1)+1)} = \frac{2}{(k+1)(k+2)}$.

Find Pk+1 for the Given Pk Pk=2k(k+1)P _ { k } = \frac { 2 } { k ( k + 1 ) }Pk​=k(k+1)2​

Find Pk₊₁ for the Given P_k $P _ { k } = \frac { 2 } { k ( k + 1 ) }$