0
$\begingroup$

Is is possible to prove that

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

Thanks in advance.

$\endgroup$
4
  • 1
    $\begingroup$ What is $\;10_{c_i}\;$ ?? $\endgroup$ – Timbuc Jan 8 '15 at 16:13
  • $\begingroup$ Oh, rats! Again, I'd love to be mind reader. $\endgroup$ – Timbuc Jan 8 '15 at 16:13
  • 4
    $\begingroup$ 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 $n=10$. $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 16:13
  • 1
    $\begingroup$ For the particular case $n=10$, an explicit numerical computation does it: a computation is a proof. $\endgroup$ – André Nicolas Jan 8 '15 at 16:51
5
$\begingroup$

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 $2$: $$ 2^9=\binom{10}{1}+\binom{10}{3}+\binom{10}{5}+\binom{10}{7}+\binom{10}{9} $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.