# Binomial Theorem. Can this be proved or not . [duplicate]

Is is possible to prove that

$${10 \choose 1 } + {10 \choose 3} + {10 \choose 5} + {10 \choose 7} + {10 \choose 9} ={2^{10-1}}$$

• What is $\;10_{c_i}\;$ ?? – Timbuc Jan 8 '15 at 16:13
• Bad formatted and a duplicate. Notice that: $$\sum_{k=0}^{n}\binom{n}{k}(-1)^k = 0,\qquad \sum_{k=0}^{n}\binom{n}{k} = 2^n.$$ Consider half the difference and plug in $n=10$. – Jack D'Aurizio Jan 8 '15 at 16:13
• For the particular case $n=10$, an explicit numerical computation does it: a computation is a proof. – André Nicolas Jan 8 '15 at 16:51
By the Binomial Theorem \begin{align} &(1+1)^{10}\\ &=\small\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}\\ &(1-1)^{10}\\ &=\small\binom{10}{0}-\binom{10}{1}+\binom{10}{2}-\binom{10}{3}+\binom{10}{4}-\binom{10}{5}+\binom{10}{6}-\binom{10}{7}+\binom{10}{8}-\binom{10}{9}+\binom{10}{10} \end{align} Add and divide by $2$: $$2^9=\binom{10}{0}+\binom{10}{2}+\binom{10}{4}+\binom{10}{6}+\binom{10}{8}+\binom{10}{10}$$ Subtract and divide by $2$: $$2^9=\binom{10}{1}+\binom{10}{3}+\binom{10}{5}+\binom{10}{7}+\binom{10}{9}$$