I-compute-k-0-1000-C-2000-2k-

Question Number 136098 by SOMEDAVONG last updated on 18/Mar/21

I) . compute \underset{k = 0}{\sum^{1000}} C_{2000}^{2 k} .

Answered by Olaf last updated on 18/Mar/21

{(1 - 1)}^{2000} = 0 = \underset{p = 0}{\sum^{2000}} {(- 1)}^{p} C_{p}^{2000}

0 = \underset{k = 0}{\sum^{1000}} {(- 1)}^{2 k} C_{2 k}^{2000} + \underset{k = 0}{\sum^{999}} {(- 1)}^{2 k + 1} C_{2 k + 1}^{2000}

0 = \underset{k = 0}{\sum^{1000}} C_{2 k}^{2000} - \underset{k = 0}{\sum^{999}} C_{2 k + 1}^{2000}

\underset{k = 0}{\sum^{1000}} C_{2 k}^{2000} = \underset{k = 0}{\sum^{999}} C_{2 k + 1}^{2000}

{(1 + 1)}^{2000} = 2^{2000} = \underset{p = 0}{\sum^{2000}} C_{p}^{2000}

2^{2000} = \underset{k = 0}{\sum^{1000}} C_{2 k}^{2000} + \underset{k = 0}{\sum^{999}} C_{2 k + 1}^{2000}

2^{2000} = \underset{k = 0}{\sum^{1000}} C_{2 k}^{2000} + \underset{k = 0}{\sum^{1000}} C_{2 k}^{2000}

2^{2000} = 2 \underset{k = 0}{\sum^{1000}} C_{2 k}^{2000}

\underset{k = 0}{\sum^{1000}} C_{2 k}^{2000} = \frac{1}{2} (2^{2000}) = 2^{1999}

Commented by SOMEDAVONG last updated on 18/Mar/21

THANKS TEACHER