I am seeking assistance in calculating the sum of a specific sequence of binomial coefficients:
$1000 \choose 1$ + $1000 \choose 5$ + $1000 \choose 9$ + … + $1000 \choose 997$.
I have noticed that the terms in the sum follow the pattern $\sum_{k=0}^{249} \binom{1000}{4k+1}$. However, I am uncertain about the next steps in simplifying this expression. I am wondering if using the properties of binomial coefficients could be helpful. I have also attempted to use Vandermonde's identity, but I haven't been able to simplify the expression significantly.
Could someone provide guidance on which properties of binomial coefficients to apply or suggest an approach to calculate the sum effectively?
Best Answer
Notice that ${1000\choose997} = {1000\choose3}$. You can rewrite your sum as
$$\sum_{k=0}^{249} {1000\choose 2k+1}$$
Which is half of the odd terms of that row of Pascal's triangle. There are standard exercises to show that the sum of the all the odd terms is $2^{999}$, so your sum should be half of that $2^{998}.$