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Use the binomial theorem to compute: $$ \sum_{i=0}^{27} \binom{27}{i} (-3)^{2i+1}$$

This is how far I got:

$$ (x+1)^{27} = \sum_{i=0}^{27} \binom{27}{i} x^i\\ -3 \cdot (-3)^{2i} = (-3)^{2i+1}\\ -3 \sum_{i=0}^{27} \binom{27}{i} (-3)^{2i}$$

I'm stuck here and don't know what to do next; could someone point me in the right direction?

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    $\begingroup$ Can you write $(-3)^{2i}$ in a way that makes the exponent be $i$? $\endgroup$ – Ian Aug 9 '17 at 9:51
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$$ \sum_{i=0}^{27} \binom{27}{i} (\color{blue}{-3})^{\color{blue}2i}= \sum_{i=0}^{27} \binom{27}{i} (\color{blue}9)^{i}=(1+9)^{27}=10^{27}$$

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