Mathematical induction?


Let Sn= 1/1x3 + 1/3x5 +1/5x7 + ....... + 1/(2n-1)(2n+1) for integers n≥1.

Prove by induction that Sn=(1/2)(1-1/(2n+1) )

I am a little confused about induction. I have done the first step as shown below.

Step 1: Prove true for n=1


RHS=1/2(1-1/2(1)+1)=1/2(1-1/3)= 1/2 x 2/3= 1/3

LHS=RHS. Therefore, n=1 is true.

Can someone please explain to me the rest of the proof in steps? Any help will be greatly appreciated. Thank you :)

1 Answer

  • 8 years ago
    Best Answer

    Generally, the next step in the proof by induction is to show that if the proposition (i.e. formula) is true for some integer, n, then it is true for n+1. In practice, this means you take as a given the proposition in n as a starting point, and try to manipulate it to show that the n+1 case is also true. In symbols, you're trying to show P(n) => P(n+1).

    When you stop and think about it, what's happening here is you're proving that whenever a proposition is true for a given number, it's also true for the next number.

    When you combine this with the fact you have proved it for n=1, you have demonstrated it is true for all natural numbers. That's because you've got the case of n=1, and your induction proof says that whenever you have a true instance of proposition for a given number, it's also true for the next one. So since it's true for n=1, it must be true for n=2, also. But if it's true for n=2, then your induction proof in n says it must be true for n=3. And so on.

    FYI, there's also nothing special about n=1. If instead, say you only were able to prove the proposition in the case of n = 8. If you're able to prove the induction step, then the fact it's true with n=8 means it's true for n=9. And because it's true for n=9, it's also true for n=10. And so on. You would have proved that the proposition (i.e. formula) holds for all numbers n, such that n greater than or equal to 8.

    If you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

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