Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)

In summary, the steps of this proof technique are as follows: step 0, verifying the base case; step 1, verifying the statement holds for all $n \ge n_0$ where $n_0$ is an integer; and induction step, which is the sum on the left-hand side of the equation above.
  • #1
simcan18
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  • #2
I know this is a cliche, but you should really show your work (also in http://mathhelpboards.com/pre-calculus-21/prove-induction-25021.html), not just because it makes others happy, but mostly because it is not useful, nor fun, to be spoon-fed an induction proof.

You are probably aware of the general structure of this proof technique, but in what is perhaps its most elementary form it looks like this. You want to prove a statement for all $n \ge n_0$ where $n_0$ is an integer:

step 0. You verify the base case, i.e. you verify the statement holds for $n = n_0$. Often, but not always, $n_0 = 0$ or $n_0 = 1$.
step 1. You verify: If the statement holds for all $n = n_0,\ldots,m$ where $m \ge n_0$ is a certain integer, then the statement also holds for $n = m + 1$.

Step 0 cannot be omitted, although it is often tempting to do so.

Now, in these two threads, why don't you start by verifying step 0. If that checks out, then formulate the induction hypotheses. (This hypothesis is the part between if and then in step 1.) Finally, take all the time that is required to perform the induction step and do not give up on it too easily.
 
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  • #3
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.
 
  • #4
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
 
  • #5
Okay, once you've shown the base case is true, you want to decide what your inductive step is going to be, given the hypothesis \(P_n\):

\(\displaystyle \sum_{k=1}^n\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}\)

I see that in order for the sum on the LHS to go up to \(n+1\), we need to add:

\(\displaystyle \frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{1}{(2n+1)(2n+3)}\)

And this gives us:

\(\displaystyle \sum_{k=1}^n\left(\frac{1}{(2k-1)(2k+1)}\right)+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\)

Or:

\(\displaystyle \sum_{k=1}^{n+1}\left(\frac{1}{(2k-1)(2k+1)}\right)=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\)

Can you show that the RHS is:

\(\displaystyle \frac{n+1}{2(n+1)+1}\) ?

If you can, then you will have derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction. :)
 
  • #6
Ok. I'll work on this some more to see if I can get it.

Thanks
 

Related to Prove by mathematical induction Σ(1/[(2k-1)(2k+1)]=n/(2n+1)

1. What is mathematical induction?

Mathematical induction is a method of mathematical proof used to prove that a statement is true for all values of a specific mathematical variable. It involves proving that the statement is true for the first value of the variable, and then showing that if the statement is true for any arbitrary value, it must also be true for the next value. This process is repeated until it can be shown that the statement is true for all values of the variable.

2. What is the formula being proven by mathematical induction in this question?

The formula being proven is Σ(1/[(2k-1)(2k+1)]=n/(2n+1), where Σ represents the summation symbol, k is the variable, and n is the number of terms in the series.

3. How is mathematical induction used to prove this formula?

First, the formula is shown to be true for the first value of k (usually k = 1). Then, it is assumed to be true for an arbitrary value of k (called the "inductive hypothesis"). Using this assumption, the formula is then shown to be true for the next value of k. This process is repeated until it is shown that the formula is true for all values of k.

4. Why is mathematical induction a valid method of proof?

Mathematical induction is considered a valid method of proof because it is based on the fundamental mathematical principle that if a statement is true for a starting value and can be shown to be true for the next value based on the previous value, then it must be true for all values. This principle is known as the principle of mathematical induction.

5. How can I use mathematical induction to solve problems?

Mathematical induction can be used to solve problems that involve proving a statement for all values of a specific variable. This can include problems related to sequences, series, and equations. To use mathematical induction, you will need to identify the starting value of the variable, the inductive hypothesis, and the method for showing that the statement is true for the next value of the variable. With practice, you can become proficient in using mathematical induction to solve various mathematical problems.

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