I am trying to write $$\frac1{\sqrt5}\left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right] $$ as a single finite series of the form $\sum^{10}_{j=0}a_j$, where $a_j$ depends on $j$.
So far I have expanded the term $(\frac{\sqrt5}2+\frac12)^{10}$ and the term $(-\frac{\sqrt5}2+\frac12)^{10}$ using the binomial theorem.
For the term $(\frac{\sqrt5}2+\frac12)^{10}$, I got $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(\frac{\sqrt5}2)^{10-m}$.
For the term $(-\frac{\sqrt5}2+\frac12)^{10}$, I got $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(-\frac{\sqrt5}2)^{10-m}$.
I am not sure how to combine the terms to write them under one series.
This is what I have written so far: $$\frac1{\sqrt5}\left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right] $$
$=$$\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(\frac{\sqrt5}2)^{10-m}$$ - $ $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(-\frac{\sqrt5}2)^{10-m}$
$=$ $\sum _{m=0}^{10}\binom{10}{m}\:\left(\left(\frac{\sqrt{5}}{2}\right)^m\left(\frac{1}{2}\right)^{10-m}-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\left(\frac{1}{2}\right)^{10-m}\right)\right)$
$=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$
The goal of the excersise is to show that $\sum^{10}_{j=0}a_j$ will give the same terms as $\frac1{\sqrt5} = \left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right]\\$
which will give same terms as $\sum^5_{j=1}\binom{10}{ 2j-1}\frac{5^{j-1}}{2^9} $
So I need help in writting $\sum^{10}_{j=0}a_j$, and so far I have $\sum^{10}_{j=0}a_j$
$=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$
The goal of the final exercise is to show that these terms yield the same terms as the series $\sum^5_{j=1}\binom{10}{2j-1}\frac{5^{j-1}}{2^9}$. Or in other words, show that
$$\frac1{\sqrt5} \left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right]\\ = \sum^5_{j=1}\binom{10}{ 2j-1}\frac{5^{j-1}}{2^9} $$
To do so, I need help in combining the expansion of the terms under one series in the form $\sum^{10}_{j=0} a_j$, where $a_j$ depends on $j$.
I mainly need someone to check if $\sum^{10}_{j=0}a_j$ $=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$ is correct or not. I can expand the terms myself and I know that they cancel out. I am also not supposed to use any calculator computations, so I am not looking for the sum of the series, just the series itself.
Thank you for any solutions and answers or any sort of help provided.
It’s not so bad if you do the simplifications in a better order:
$$\begin{align*} &\frac1{\sqrt5}\left(\left(\frac{\sqrt5}2+\frac12\right)^{10}-\left(-\frac{\sqrt5}2+\frac12\right)^{10}\right)\\ &\qquad=\frac1{2^{10}\sqrt5}\left(\sum_{k=0}^{10}\binom{10}k5^{k/2}-\sum_{k=0}^{10}(-1)^k\binom{10}k5^{k/2}\right)\\ &\qquad=\frac1{2^{10}\sqrt5}\sum_{k=0}^{10}\binom{10}k5^{k/2}\left(1-(-1)^k\right)\\ &\qquad=\frac1{2^9\sqrt5}\sum_{k=0}^4\binom{10}{2k-1}5^{k+\frac12}\\ &\qquad=\frac1{2^9}\sum_{k=0}^4\binom{10}{2k+1}5^k\\ &\qquad=\frac1{2^9}\sum_{k=1}^5\binom{10}{2k-1}5^{k-1} \end{align*}$$
The first step factors out the denominator and applies the binomial theorem to $\left(\sqrt5+1\right)^{10}$ and $\left(-\sqrt5+1\right)^{10}$; the third recognizes that the terms with even $k$ cancel out and keeps only the terms with odd $k$; and the last simply shifts the index by $1$.