Recent content by simcan18
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Graduate Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)
Ok. I'll work on this some more to see if I can get it. Thanks- simcan18
- Post #6
- Forum: General Math
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Graduate Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)
So, do I start like this n=1 1 / (2x1-1) x(2x1+1) = 1/ 2(1) + 1 = 1/3, so both equations are true for n=1, as it is 1/3- simcan18
- Post #4
- Forum: General Math
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MHB Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7
I have done the base case and some of the inductive..which I'm not sure I'm going in the right direction. Inductive, So does it hold true for n=k+1 3 raised 2(k+1)+1 +2 raised(k+1)-1 = 3 raised 2k+2+1 +2 raised (k+1)-1 = 3 raised 2k+1 x 3 raised2 + 2 raised k x 2 raised 0 =9 x 3 raised...- simcan18
- Post #2
- Forum: General Math
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Graduate Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)
Thanks, but I'm unsure of how to get started with the problem as I don't understand which is the reason for no work being shown.- simcan18
- Post #3
- Forum: General Math
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Graduate Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)
I'm in need of assistance for the following attachement.- simcan18
- Thread
- Induction Mathematical Mathematical induction
- Replies: 5
- Forum: General Math
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MHB Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7
Can someone with understanding of proof by induction help with this problem? Prove by induction that 3 raised to 2n+1 + 2 raised to n-1 is divisible by 7 for all numbers greater than/or equal to 1. How do you do the inductive step?- simcan18
- Thread
- Induction
- Replies: 2
- Forum: General Math