№ 20.3 Алгебра = № 37.3 Математика
Знайдіть суму шести перших членів геометричної прогресії $(b_{n})$, якщо:
1) $b_{1}=81$, $q=\frac{1}{3}$;
2) $b_{1}=-17$, $q=-2$;
3) $b_{1}=32$, $q=-\frac{1}{2}$;
4) $b_{1}=5\sqrt{2}$, $q=\sqrt{2}$.
Розв’язок:
1) $S_{6}=\frac{81\cdot(1-(\frac{1}{3})^{6})}{1-\frac{1}{3}}=\frac{81\cdot(1-\frac{1}{729})}{\frac{2}{3}}=$
$=\frac{81\cdot\frac{728}{729}}{\frac{2}{3}}=\frac{\frac{728}{9}}{\frac{2}{3}}=\frac{728\cdot3}{9\cdot2}=$
$=\frac{364}{3}=121\frac{1}{3}$
2) $S_{6}=\frac{-17\cdot(1-(-2)^{6})}{1-(-2)}=\frac{-17\cdot(1-64)}{3}=$
$=\frac{-17\cdot(-63)}{3}=-17\cdot(-21)=357$
3) $S_{6}=\frac{32\cdot(1-(-\frac{1}{2})^{6})}{1-(-\frac{1}{2})}=\frac{32\cdot(1-\frac{1}{64})}{\frac{3}{2}}=$
$=\frac{32\cdot\frac{63}{64}}{\frac{3}{2}}=\frac{\frac{63}{2}}{\frac{3}{2}}=\frac{63}{2}\cdot\frac{2}{3}=21$
4) $S_{6}=\frac{5\sqrt{2}\cdot(1-(\sqrt{2})^{6})}{1-\sqrt{2}}=\frac{5\sqrt{2}\cdot(1-8)}{1-\sqrt{2}}=$
$=\frac{-35\sqrt{2}}{1-\sqrt{2}}=\frac{-35\sqrt{2}\left( 1+\sqrt{2} \right)}{\left( 1-\sqrt{2} \right)\left( 1+\sqrt{2} \right)}=$
$=\frac{-35\sqrt{2}-35\cdot2}{1-2}=$
$=\frac{-35\sqrt{2}-70}{-1}=35\sqrt{2}+70$