#1 If you meant
Pk=2^(k-1)/k!
then Pk+1 = 2^k/(k+1)!
#2 If
Pk = 7 + 13 + 19 + ...+[6(k - 1)+1] + (6k + 1)
Pk+1 = 7+13+...+(6k+1)+(6(k+1)+1)
= 7+13+...+(6k+1)+(6k+7)
#3 The first step is to show that P(1) is true
#4 You assume that
3 + 11 + 19 + 27 + … + (8n - 5) = n(4n - 1)
Then the next step is to add the same term to both sides:
3 + 11 + 19 + 27 + … + (8n-5) + (8(n+1)-5) = n(4n-1)+(8(n+1)-5)
Then you have to show that
n(4n-1)+(8(n+1)-5) = (n+1)(4(n+1)-1)
Can you please check my answers?
1.Find Pk + 1 if Pk=2^K-1/k!
answer: 2^k+1/(k+1)!
2.Find Pk + 1 if Pk = 7 + 13 + 19 + ...+[6(k - 1)+1] + (6k + 1)
answer: 7+13+9...(6k-1+1)+6k+1 +(6k+2)
3.What is the first step when writing a proof using mathematical induction?
answer: Find the sum of the integers
4.Which of the following shows the correct first step to prove the following by mathematical induction?
3 + 11 + 19 + 27 + … + (8n - 5) = n(4n - 1)
answer: 3+11+19+27+ + (8k-5)+8(k+1)-5 = (k+1)(4(k+1)-1)
2 answers
I think this link nicely fits with your problem, carefully study the 3 steps I used to do an induction problem. The problem is the same as your #4
http://www.jiskha.com/display.cgi?id=1481512314
#1. I don't like your notation, did you mean
P(k+1) and P(k) ?
or Pk+1 and Pk
if Pk = 2^K - 1/k!
then Pk+1 = 2^(k+1) - 1/(k+1)!
I just replaced the k with k+1
I disagree with the answer you gave.
Why did the subtraction change to addition?
http://www.jiskha.com/display.cgi?id=1481512314
#1. I don't like your notation, did you mean
P(k+1) and P(k) ?
or Pk+1 and Pk
if Pk = 2^K - 1/k!
then Pk+1 = 2^(k+1) - 1/(k+1)!
I just replaced the k with k+1
I disagree with the answer you gave.
Why did the subtraction change to addition?
2 answers