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Need mathematical Inudction Help (1 Viewer)
- Thread starter blazer78
- Start date
darkliight
I ponder, weak and weary
a) What can't you do with this one? Give me a start and i'll help you finish. 
b) After you prove it true for n = 1
Assume true for n = k: 1/2! + 2/3! + 3/4! + ... + k/(k+1)! = ((k+1)! - 1) / (k+1)!
Prove true for n = k + 1:
1/2! + 2/3! + 3/4! + ... + k/(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
((k+1)! - 1) / (k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
(k+2)((k+1)! - 1) / (k+2)(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)! **multiplying the first fraction by (k+2) / (k+2)**
((k+2)! - k - 2)) / (k+2)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
((k+2)! - 1) / (k+2)! = ((k+2)! - 1) / (k+2)!
b) After you prove it true for n = 1
Assume true for n = k: 1/2! + 2/3! + 3/4! + ... + k/(k+1)! = ((k+1)! - 1) / (k+1)!
Prove true for n = k + 1:
1/2! + 2/3! + 3/4! + ... + k/(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
((k+1)! - 1) / (k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
(k+2)((k+1)! - 1) / (k+2)(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)! **multiplying the first fraction by (k+2) / (k+2)**
((k+2)! - k - 2)) / (k+2)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
((k+2)! - 1) / (k+2)! = ((k+2)! - 1) / (k+2)!
Here's my solution to the main proof in 3 b) :
http://img114.imageshack.us/my.php?image=induction8us.png
:http://img114.imageshack.us/my.php?image=induction8us.png
darkliight
I ponder, weak and weary
Headed to bed now, but here is a good start to your first problem.
After we know it works for n = 1, we assume it's true for k: 7^k - 3^k = 4A (A is an integer)
We need to show it's true for k + 1:
7^(k+1) - 3^(k+1)
7*7^(k) - 3*3^(k)
(4+3)*7^(k) - 3*3^(k)
4*7^(k) + 3*7^(k) - 3*3^(k)
It's all your to finish off, keeping in mind that 7^k - 3^k = 4A
After we know it works for n = 1, we assume it's true for k: 7^k - 3^k = 4A (A is an integer)
We need to show it's true for k + 1:
7^(k+1) - 3^(k+1)
7*7^(k) - 3*3^(k)
(4+3)*7^(k) - 3*3^(k)
4*7^(k) + 3*7^(k) - 3*3^(k)
It's all your to finish off, keeping in mind that 7^k - 3^k = 4A
darkliight
I ponder, weak and weary
Line three:
(k+2)((k+1)! - 1) / (k+2)(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
The only part that changed, from this to line 4, was the first fraction: (k+2)((k+1)! - 1) / (k+2)(k+1)!
because I multiplied it by (k+2)/(k+2), I just simplified it.
So the numerator is now:
(k+2)((k+1)! - 1) = (k+2)(k+1)! - 1(k+2) = (k+2)! - k - 2
The denominator now is:
(k+2)(k+1)! = (k+2)!
The main thing here was knowing that, in general, (n+1)*n! = (n+1)!
so, for example, 3*2! = 3*2*1 = 3!
and in our case, (k+2)(k+1)! = (k+2)!
Hope that helps,
cheers
(k+2)((k+1)! - 1) / (k+2)(k+1)! + (k+1) / (k+2)! = ((k+2)! - 1) / (k+2)!
The only part that changed, from this to line 4, was the first fraction: (k+2)((k+1)! - 1) / (k+2)(k+1)!
because I multiplied it by (k+2)/(k+2), I just simplified it.
So the numerator is now:
(k+2)((k+1)! - 1) = (k+2)(k+1)! - 1(k+2) = (k+2)! - k - 2
The denominator now is:
(k+2)(k+1)! = (k+2)!
The main thing here was knowing that, in general, (n+1)*n! = (n+1)!
so, for example, 3*2! = 3*2*1 = 3!
and in our case, (k+2)(k+1)! = (k+2)!
Hope that helps,
cheers
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