Find pk_{+ 1} for the given pk. $p _ { k } = \frac { k ^ { 2 } } { 7 ( k + 2 ) ^ { 2 } }$ A) $p _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 3 ( k + 8 ) ^ { 2 } }$ B) $p _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 8 ( k + 8 ) ^ { 2 } }$ C) $p _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 8 ( k + 3 ) ^ { 2 } }$ D) $p _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 7 ( k + 3 ) ^ { 2 } }$ E) $p _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 7 ( k + 8 ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

The correct expression for $p_{k+1}$ is obtained by substituting $k+1$ for $k$ in the given formula for $p_k$, resulting in $p_{k+1} = \frac{(k+1)^2}{7(k+3)^2}$.

Find Pk + 1 for the Given Pk pk=k27(k+2)2p _ { k } = \frac { k ^ { 2 } } { 7 ( k + 2 ) ^ { 2 } }pk​=7(k+2)2k2​

Find Pk_{+ 1} for the Given Pk $p _ { k } = \frac { k ^ { 2 } } { 7 ( k + 2 ) ^ { 2 } }$