Find Pk+1 for the given Pk. $P _ { k } = \frac { k ^ { 2 } } { 6 } ( 5 k + 5 )$
A) $P _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 6 } ( 5 k + 10 )$
B) $P _ { k + 1 } = \frac { k ^ { 2 } } { 6 } ( 5 k + 5 ) + 1$
C) $P _ { k + 1 } = \frac { k ^ { 2 } } { 6 } ( 5 k + 10 )$
D) $P _ { k + 1 } = \frac { k ^ { 2 } + 1 } { 6 } ( 5 k + 10 )$
E) $P _ { k + 1 } = \frac { ( k + 1 ) ^ { 2 } } { 6 } ( 5 k + 5 )$

Question

Quizplus · Accepted Answer

To find $P_{k+1}$, substitute $k+1$ into the original formula: $P_{k+1} = \frac{(k+1)^2}{6}(5(k+1) + 5)$. Simplifying inside the parentheses gives $5k + 10$, matching choice A.

Find Pk+1 for the Given Pk Pk=k26(5k+5)P _ { k } = \frac { k ^ { 2 } } { 6 } ( 5 k + 5 )Pk​=6k2​(5k+5)

Find Pk+1 for the Given Pk $P _ { k } = \frac { k ^ { 2 } } { 6 } ( 5 k + 5 )$