# Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$

I have a few students that are having trouble understanding that $$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$ specifically that $$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} = 2n+1.$$ I've tried explaining it a few different ways, but it didn't seem to go over very well. What would be a good way to explain this?

• Perhaps demonstrate it for low values of $n$, and then sneak in an induction proof? Commented Mar 2, 2015 at 1:51
• From what you wrote, one thing that I notice that the students would find confusing is that the products end at different values. Perhaps write out the product (or have the students figure out) the last three factors of both the numerator and denominator. Then, they can see some cancellation. If this is an intro to proof class, they might appreciate learning product notation at this point; with product notation, it becomes somewhat clearer. Commented Mar 2, 2015 at 1:56
• Maybe put $x_n=1·3·5···(2n-1)$ and write $\frac{x_n(2n+1)}{x_n}$ Commented Mar 2, 2015 at 1:58
• Perhaps point out that $$2n-1= 2(n-1)+1$$ $$2n-3=2(n-2)+1$$ etc.. Commented Mar 2, 2015 at 14:46

I would suggest including more terms at the end of the products. That makes it easier to see the pattern and to see why it's just $(2n+1)$ that's left at the end.

\require{cancel} \begin{align} \frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} &= \frac{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)(2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)\quad\quad\quad} \\ &=\frac{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}(2n+1)}{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}\quad\quad\quad} \\ &=2n+1 \end{align}

It might also simply be a matter of spacing: students might get confused by seeing the $2n+1$ directly above the $2n-1$. I've tried to format the above to emphasize that the $2n+1$ is an extra factor that doesn't appear in the bottom.

Finally, it could also be helpful to pick a small $n$ and work out the product concretely.

I would suggest put some various number for example n=3 , and force them to write numbers in two rows then simplify $$n=3 \to 2n+1=2(3)+1=7 \to1.3.5.7 \\2n-1=2(3)-1=5 \to 1.3.5\\ \frac{1.3.5...(2n+1)}{1.3.5...(2n-1)}=\frac{1.3.5.7}{1.3.5}$$

It might have helped if you did the equation in reverse, but it seems like they're not seeing the general pattern, namely, that the $j$th factor in $1 \cdot 3\cdots \cdot (2n + 1)$ is $2j - 1$. If you could explain that, then they'll understand that the $n$th factor of the sequence is $2n - 1$, and the $n+1$st factor is $2(n+1) - 1 = 2n + 1$.

Just give an example with a sufficiently large value of $n$, nothing works better.

Suppose $n=10$, then:

• $1\cdot3\cdot5\cdot\ldots\cdot(2n+1)=$
• $1\cdot3\cdot5\cdot\ldots\cdot21=$
• $1\cdot3\cdot5\cdot\ldots\cdot19\cdot21=$
• $1\cdot3\cdot5\cdot\ldots\cdot(2n-1)\cdot(2n+1)$

Numeric examples are also very useful when approaching series/sequences related problem, as they allow you to understand the behavior of the expression at hand before trying to prove it.