What is the general term for $1+1\cdot3+1\cdot3\cdot5+1\cdot3\cdot5\cdot7$?"

Question

$$1+1\cdot3+1\cdot3\cdot5+1\cdot3\cdot5\cdot7$$

How to find an expression for the general term of a series ?

Do you mean a general term, or a general partial sum? In other words, do you want a nice expression for $1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11$, or a nice expression for $$1+1\cdot 3 + 1\cdot 3\cdot 5 + \cdots + 1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11$$(I don't think the latter one has a nice expression, but that's a different issue.) — Arthur, Mar 03 '20 at 15:20
$$\frac{(2n)!}{2^n\cdot n!}$$ is the expression $1\cdot 3\cdot 5\cdot \cdots\cdot (2n-1).$ — Thomas Andrews, Mar 03 '20 at 15:22
Your partial sums is OEIS sequence A099953 which is one less than OEIS sequence A076795. There is no closed form known. — Somos, Mar 03 '20 at 16:19
No idea what you mean by "3rd variable," but the third term in the sum is $1\cdot 3\cdot 5=15.$ @MaximPlaton — Thomas Andrews, Mar 03 '20 at 17:03

score 2 · Answer 1 · edited Mar 03 '20 at 15:49

2

The equation for the product of consecutive odd numbers from $1$ to $2n-1$ can be given as $$a_n=\frac{(2n)!}{2^n n!}$$

Therefore the total sum can be written as $$\sum_{i=1}^na_i = \sum_{i=1}^n\frac{(2i)!}{2^i i!}$$

edited Mar 03 '20 at 15:49

2'5 9'2

answered Mar 03 '20 at 15:23

lioness99a

I think you mean product, not sum. – marty cohen Mar 03 '20 at 15:39
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@martycohen The first is a product (multiply) and the second is a sum (add) – lioness99a Mar 03 '20 at 15:40
@lioness99a, thank you is right – Maxim Platon Mar 03 '20 at 17:34

Boy · Answer 2 · 2020-04-18T18:00:35.093

You can try this also:

Start with a product series ending with an odd number:- $$1\cdot 2 \cdot 3 \cdot … \cdot 2r-1$$

Now that we only want the odd numbers, divide all of the even ones by 2. There are 'r' odd numbers and 'r-1' even numbers.

$$\frac{1\cdot 2 \cdot 3 \cdot … \cdot 2r-1}{2^{r-1}}$$

Now we are left with:

$$1\cdot 1 \cdot 3 \cdot 2 \cdot… \cdot 2r-1$$

Factoring:

$$(1\cdot 2 \cdot 3 \cdot … \cdot r-1)\cdot 1 \cdot 3 \cdot 5\cdot … \cdot 2r-1$$

Divide by $r-1!$ and we are left with:

$$1 \cdot 3 \cdot 5\cdot … \cdot 2r-1$$

Thus our general term is: $$t_r=\frac{(2r-1)!}{2^{r-1}(r-1)!}$$

2 Answers2