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.
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1What is $;10_{c_i};$ ?? – Timbuc Jan 08 '15 at 16:13
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Oh, rats! Again, I'd love to be mind reader. – Timbuc Jan 08 '15 at 16:13
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4Bad 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 08 '15 at 16:13
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1For the particular case $n=10$, an explicit numerical computation does it: a computation is a proof. – André Nicolas Jan 08 '15 at 16:51
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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 $2$: $$ 2^9=\binom{10}{1}+\binom{10}{3}+\binom{10}{5}+\binom{10}{7}+\binom{10}{9} $$
robjohn
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