<s>[INST] Is there any way to calculate $\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$ faster? During solving a problem I get this expression:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$$
To calculate it normally, I do this way:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}=2\binom{10}{1}-2\binom{10}{3}+\binom{10}{5}$$
And here we have to evaluate $\binom{10}{3}$ and $\binom{10}{5}$. after all I got $32$ as the result correctly.
My question: Is there any other way to calculate this expression easier (faster)? I suspect there is because the answer is $32=2^5$ and I guess maybe there is other way to calculate it for example by combinations or other ways.
 [/INST] $$(1+i)^{10} = {10\choose 0} + {10\choose 1}i -{10\choose 2}-{10\choose 3}i +{10\choose 4}+ ... -{10\choose 10} $$
$$(1-i)^{10} = {10\choose 0} - {10\choose 1}i -{10\choose 2}+{10\choose 3}i +{10\choose 4}+...-{10\choose 10}$$
So $$...={(1+i)^{10}-(1-i)^{10}\over 2i} = {1\over 2i}((2i)^5-(-2i)^5) = 32$$
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