# Thread: Combining two series (Binomial Expansion)

1. ## Combining two series (Binomial Expansion)

I've been working through this binomial question in my notes and there is one step I'm not entirely sure about. When it comes to the point of combining the two series into one I'm fairly certain what's in my notes is incorrect so I'd like to fix it.

It's safe to assume that all the work up to this point is correct. (See figure)

How do I combine these two series into 1?

I think my professor tried to do something like a common denominator but I can't follow his work.

Someone care to show me (In slow steps ) how I can combine the two and simplify?

Thanks again!

2. The reason you haven't got a reply yet might be that it isn't clear what you're doing.
It's not clear to me at least. You should probably rewrite the question using latex.

3. Originally Posted by TheCoffeeMachine
The reason you haven't got a reply yet might be that it isn't clear what you're doing.
It's not clear to me at least. You should probably rewrite the question using latex.
All I want to do is combine the two series in the 2nd last line into one. It's simply algebra issues. What are you confused about?

4. Originally Posted by jegues
All I want to do is combine the two series in the 2nd last line into one. It's simply algebra issues. What are you confused about?
Ok. I didn't see the question mark, haha.

$\displaystyle \displaystyle \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n}{3^{n-2}(n-2)!}+\sum_{n=3}^{\infty}\frac{\prod_{k=1}^{n}\left (3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle = \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)}{3^{n-2}(n-2)!3^2(n)(n-1)}+\sum_{n=3}^{\infty}\frac{\prod_{k=1}^{n}\left( 3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle = \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)}{3^nn!}+\sum_{n=3}^{\infty}\frac{\prod_{k=1}^{n }\left(3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle = \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)+\prod_{k=1}^{n}\left(3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

5. Originally Posted by TheCoffeeMachine
Ok. I didn't see the question mark, haha.

$\displaystyle \displaystyle \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n}{3^{n-2}(n-2)!}+\sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left (3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)}{3^{n-2}(n-2)!3^2(n)(n-1)}+\sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left( 3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)}{3^nn!}+\sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n }\left(3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$

$\displaystyle \displaystyle \sum_{n=3}^{\infty}\frac{\prod_{k=3}^{n}\left(3k-8\right)(-1)^{n}(x-5)^n3^2(n)(n-1)+\prod_{k=3}^{n}\left(3k-2\right)(-1)^{n+1}(x-5)^n}{3^nn!}$
I'm still a little confused about the 2nd and 3rd lines. Are you trying to find a common denominator between the two at that point?

I see one series with an $\displaystyle 3^{n-2}(n-2)!$ in the denominator and the other with $\displaystyle 3^{n}n!$,

So if I want to transform the first series to have the same denominator as the 2nd series we multiply both top and bottom by,

$\displaystyle 3^{2}(n)(n-1)$

but how does this disapeer from the denominator in the 3rd line, I thought we would have used to terms on the numerator to cancel what we didn't want on the denominator.

Can you please clarify what I'm missing or not seeing?

Thanks again!

6. Originally Posted by jegues
but how does this disapeer from the denominator in the 3rd line, I thought we would have used to terms on the numerator to cancel what we didn't want on the denominator.
$\displaystyle 3^{n-2}3^2 = 3^n$ and $\displaystyle n(n-1)(n-2)! = n!$, so $\displaystyle 3^{n-2}(n-2)!3^2(n)(n-1) = 3^nn!$

7. ## Combining two series

Originally Posted by jegues
I've been working through this binomial question in my notes and there is one step I'm not entirely sure about. When it comes to the point of combining the two series into one I'm fairly certain what's in my notes is incorrect so I'd like to fix it.

It's safe to assume that all the work up to this point is correct. (See figure)

How do I combine these two series into 1?

I think my professor tried to do something like a common denominator but I can't follow his work.

Someone care to show me (In slow steps ) how I can combine the two and simplify?

Thanks again!

First, I'll put your next to last line in LaTeX. (Well, I combined the terms that were outside the summations.)

$\displaystyle \displaystyle f(x)={7\over 9}(x-5)^2+{{(x-5)\over 3}-1 \\$

$\displaystyle \displaystyle + \sum_{n=3}^{\infty}{{1\cdot 4\cdot 7 \dots (3n-8)(-1)^n(x-5)^n}\over{3^{(n-2)}(n-2)!}}$

$\displaystyle \displaystyle +\sum_{n=3}^{\infty}{{1\cdot 4\cdot 7 \dots (3n-2)(-1)^{n+1}(x-5)^n}\over{3^{n}n!}}$

Multiply the expression in the first summation by $\displaystyle \displaystyle {{9n(n-1)}\over{9n(n-1)}}$ to have a common denominator with the expression in the second summation.

in the second summation, write $\displaystyle (-1)^{n+1}$ as $\displaystyle -1\cdot (-1)^n$

and $\displaystyle \dots (3n-2)$ as $\displaystyle \dots (3n-8)\cdot (3n-5)\cdot (3n-2)$

You can then combine the summations.

The expression in the resulting summation is: $\displaystyle \displaystyle {{1\cdot 4\cdot 7 \dots (3n-8)[9n^2-9n-(9n^2-21n+10)](-1)^{n}(x-5)^n}\over{3^{n}n!}}$.

This simplifies to: $\displaystyle \displaystyle {{1\cdot 4\cdot 7 \dots (3n-8)[12n-10](-1)^{n}(x-5)^n}\over{3^{n}n!}}$.

Therefore, $\displaystyle \displaystyle f(x)={7\over 9}(x-5)^2+{{(x-5)\over 3}-1 + 2 \sum_{n=3}^{\infty}(6n-5){{1\cdot 4\cdot 7 \dots (3n-8)(-1)^{n}(x-5)^n}\over{3^{n}n!}}$.