<s>[INST] Simple Summation Proof with identities Using some of the identities, determine the value of
$\sum_0^5$ ${12 \choose i}$
. You may use the
substitution ${12 \choose 6}$
= 924, but you may not evaluate the individual chooses.
Proofs of summations are a hard topic for me, hate to show no work, but i am unsure of how to approach this question. Is there any tips as well that you can give when approaching summation problems with proofs? Thankyou. 
 [/INST] $$\begin{array}?
&& 2^{12} \\ &=& (1+1)^{12} \\&=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{7} + \binom{12}{8} + \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12} \\
&=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{5} + \binom{12}{4} + \binom{12}{3} + \binom{12}{2} + \binom{12}{1} + \binom{12}{0} \\
&=& 2\left(\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}\right) + \binom{12}{6}
\end{array}$$
</s>