Question

$a) \frac{2^{2013} }{3} - \frac{2^{2012} }{3} - \frac{2^{2011 } }{3} - ... - \frac{2}{3} = ?$
$b) \frac{ 3^{2013} }{5} - \frac{2* 3^{2012} }{5} - \frac{2* 3^{2011} }{5} - ... - \frac{2*3}{5}$

Answer 1

$a)~2^{n+1}-2^n=2^n \cdot 2-2^n=2^n(2-1)=2^n \cdot 1= 2^n. \\ \\ S= \frac{(2^{2013}-2^{2012})-2^{2011}-...-2}{3}= \\ \\ =\frac{2^{2012}-2^{2011}-...-2}{3}= \\ \\ = \frac{2^{2011}-...-2}{3}= \\ \\ =..........= \\ \\ =\frac{2^2-2}{3}= \\ \\ = \frac{2}{3}.$

$b)~3^{n+1}-2 \cdot 3^{n}=3^{n} \cdot 3-2 \cdot 3^n=3^n(3-2)=3^n \cdot 1=3^n= \\ \\ S= \frac{(3^{2013}-2 \cdot 3^{2012})-2 \cdot 3^{2011}-...-2 \cdot 3}{5}= \\ \\ = \frac{3^{2012}-2 \cdot 3^{2011}-...-2 \cdot 3}{5}= \\ \\ = \frac{3^{2011}-...-2 \cdot 3}{5}= \\ \\ = ...........=\\ \\ =\frac{3^2-2 \cdot 3}{5}= \\ \\ = \frac{3}{5}.$