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!}} $.