Sum

We can see that if we put in all of the odd terms, we get $2^{99}$ , however, the odd terms are missing. 

But, we see that for each even term $\binom{99}{2x}$ there is a odd term $\binom{99}{99-2x}$ which is equal to it.

So, the sum is $2^{98}$.

Find the value of the sum
$\dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}$.

$\text{Formula: } \qquad \dbinom{n}{k}= \dbinom{n}{n-k}$

$\begin{array}{|rcll|} \hline && \mathbf{ \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{96}+ \dbinom{99}{98}} \\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4} + \dots + \dbinom{99}{48} \\ && +\dbinom{99}{98}+ \dbinom{99}{96}+ \dbinom{99}{94} + \dots +\dbinom{99}{50}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{2} + \dbinom{99}{4}+ \dbinom{99}{6} + \dots + \dbinom{99}{48} \\ && + \dbinom{99}{1}+ \dbinom{99}{3} + \dbinom{99}{5} + \dbinom{99}{7}+ \dots +\dbinom{99}{49}\\\\ &=& \dbinom{99}{0} + \dbinom{99}{1} + \dbinom{99}{2}+ \dbinom{99}{3} + \dbinom{99}{4}+ \dots \\ && + \dbinom{99}{46} + \dbinom{99}{47} + \dbinom{99}{48}+\dbinom{99}{49}\\\\ &=& \dfrac{2^{99}}{2} \\\\ &=& \mathbf{ 2^{98} } \\ \hline \end{array}$

Thanks Ilovepizza and Heureka,

Where does 2^99 come from?  I understand everything else. 

Where does $2^{99}$ come from?

Pascal's triangle:

See the sum of the each line:
$\small{ \begin{array}{|l|rclcl|} \hline n \\ \hline 0 &\dbinom{0}{0} &=& 1 &=& 2^0 \\ 1 &\dbinom{1}{0}+\dbinom{1}{1} &=& 1+1 &=& 2^1 \\ 2 &\dbinom{2}{0}+\dbinom{2}{1}+\dbinom{2}{2} &=& 1+2+1 &=& 2^2 \\ 3 &\dbinom{3}{0}+\dbinom{3}{1}+\dbinom{3}{2}+\dbinom{3}{3} &=& 1+3+3+1 &=& 2^3 \\ 4 &\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} &=& 1+4+6+4+1 &=& 2^4 \\ 5 &\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5} &=& 1+5+10+10+5+1 &=& 2^5 \\ \ldots \\ \color{red}99 & \dbinom{99}{0} + \dbinom{99}{1} + \dots + \dbinom{99}{98}+\dbinom{99}{99} &=&1+99+\dots + 99 + 1 &=& \color{red}2^{99} \\ \hline \end{array} }$

$\begin{array}{|rcll|} \hline && \color{red}(1+1)^n \\\\ &=&\sum \limits_{k=0}^{n}\dbinom{n}{k} \\ \\ &=& \dbinom{n}{0}+\dbinom{n}{1} + \dots + \dbinom{n}{n-1}+\dbinom{n}{n} \\\\ &=& \color{red}2^n \\ \hline \end{array}$