The lesson that I am working on is called mathematical induction. The only problem is that the lesson was not explained very clearly due to a problem with the recording so I have no idea how to complete this problem.
Find p k+1(This is little under the p) for the given Pk(K is little).
Pk = k/2(3k-1)
Thank you for helping!!
for Pk+1 put k+1 in for k in each instance.
Pk+1 = (k+1)/2(3k+3-1)
then simplify
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To find the expression for Pk+1, we need to understand the concept of mathematical induction first. Mathematical induction is a method used to prove a statement is true for all positive integers. It typically involves two steps: the base case and the inductive step.
1. Base Case: Show that the statement is true for a specific integer, usually the smallest possible value. In this case, we can start with k = 1.
P1 = 1/2(3(1) - 1) = 1/2(3 - 1) = 1/2(2) = 1.
2. Inductive Step: Assume that the statement is true for some integer n = k, and then prove that it holds for n = k + 1.
Assume Pk is true: Pk = k/2(3k - 1).
Now, we want to find Pk+1. To do that, we need to substitute k + 1 for k in the expression for Pk.
Pk+1 = (k + 1)/2(3(k + 1) - 1).
Next, we simplify the expression:
Pk+1 = (k + 1)/2(3k + 3 - 1).
Pk+1 = (k + 1)/2(3k + 2).
Pk+1 = (k + 1)/2(3k + 2).
Therefore, the expression for Pk+1 is (k + 1)/2(3k + 2).