MCQ
$\sum_{r=0}^n 4^r \cdot{ }^n C_r$ is equal to
- A$6^{ n }$
- B$5^{-n}$
- C$4^{ n }$
- ✓$5^{ n }$
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$\frac{{\left (sin 36^o + cos 36^o - \sqrt 2 sin 27^o)( {\sin {{36}^0} + \cos {{36}^0} - \sqrt 2 \sin {{27}^0}} \right)}}{{2\sin {{54}^0}}}$ is less than