22. [-/5 Points] DETAILS HUNTERDM3 3.2.004. MY NOTES ASK YOUR TEACHER Consider the following recurrence relation: if n = 0 P ( m ) = (5 . P (n - 1 ) + 1 ifn > 0. Prove by induction that P(n) = 5" - 1 for all n 2 0. (Induction on n.) Let f(n) = 5" - 1. Base Case: If n = 0, the recurrence elation says that P(0) = 0, and the formula says that A(0) = 5 - 1 - , so they match. Inductive Hypothesis: Suppose as inductive hypothesis that P(k - 1) = for some * > 0 . Inductive Step: Using the recurrence relation, P(k) = 5 . P(k - 1) + 1, by the second part of the recurrence relation = 5 . + 1, by inductive hypothesis 1+ 4 so, by induction, P(n) = f(n) for all n 2 0. eBook 23. [1/7 Points] DETAILS PREVIOUS ANSWERS HUNTERDM3 3.2.005. MY NOTES ASK YOUR TEACHER Consider the follow currence relation: if n = 0 c(n ) = \ + 3 . c ( n - 1 ) ifn = 0 . Prove by induction that C(n) = 30 + 1 - 2n - 3 for all n 2 0. (Induction on n.) Let A(n) = 30*1 - 2n - 3 Base Case: If n = 0, the recurrence relation says that C(0) = 0, and the formula says that f(0) = X - 2 . 0 - 3 = X , so they match. Inductive Hypothesis: Suppose as inductive hypothesis that C(k - 1) = for some k > 0. Inductive Step: Using the recurrence relation, C(k) = k + 3 . C(k - 1), by the second part of the recurrence relation = k + 3 . ( ). by inductive hypothesis calcPa Operation 4k Function X Vectors so, by induction, C(n) = f(n) for all n 2 0. This eBook GREEK