Question

Cum se rezolva ex    3^0+3^1+3^2+......+3^33=?

Answer 1

$3^0\to1 \\ 3^1\to3 \\\\ S= 1+3+3^2+3^3+...+3^{33} \\\\ a_1=1 \\ a_2=1*3=3 \\ a_3=3*3=3^2 \\ a_4=3*3^2=3^3 \\..............................\\ a_{34}=3*3^{32}=3^{33} \\\\ Suma \ termenilor \ prin \ progresie \ geometica \ este \\\\ S_n=\frac{a_1-a_n*q}{1-q} \\\\ a_1\to primul \ termen \ al \ S_n \\ a_n\to ultimul \ termen \ al \ S_n \\ q\to ratia \\\\\\ S_{34}= \frac{1-3^{33}*3}{1-3}=\\\\ S_{34}= \frac{1-3^{34}}{-2} \\\\ S_{34}= \frac{3^{34}-1}{2}$