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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.

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why did you write o instead of 0 –  user58512 Jan 24 '13 at 6:41
    
My bad, it was meant to be zero. –  MatthewL Jan 24 '13 at 6:48

1 Answer 1

$$\begin{array}? && 2^{12} \\ &=& (1+1)^{12} \\&=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{7} + \binom{12}{8} + \binom{12}{9} + \binom{12}{10} + \binom{12}{11} + \binom{12}{12} \\ &=& \binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} + \binom{12}{6} + \binom{12}{5} + \binom{12}{4} + \binom{12}{3} + \binom{12}{2} + \binom{12}{1} + \binom{12}{0} \\ &=& 2\left(\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}\right) + \binom{12}{6} \end{array}$$

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(+1) Good observation. –  Mhenni Benghorbal Jan 24 '13 at 6:54

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