Question

Найти суму первых 7-ми членов геометрической прогрессии, b2=1/2, b4=1/4

Answer 1

$b_2=frac{1}{2}; ,; ; b_4=frac{1}{4}\\b_2=b_1cdot q=frac{1}{2}\\b_4=b_1cdot q^3=frac{1}{4}\\b_2cdot b_4=b_3^2; ; to ; ; b_3^2=frac{1}{2}cdot frac{1}{4}=frac{1}{8}; ,; ; b_3=frac{1}{sqrt8}=frac{1}{2sqrt2}\\q=frac{b_4}{b_3}=frac{2sqrt2}{4}=frac{sqrt2}{2}\\b_1=frac{b_2}{q}=frac{1/2}{sqrt2/2}=frac{1}{sqrt2}\\S_7=frac{b_1cdot (q^7-1)}{q-1}=frac{frac{1}{sqrt2}cdot ((frac{sqrt2}{2})^7-1)}{frac{sqrt2}{2}-1}=frac{2cdot ((sqrt2)^7-2^7)}{sqrt2cdot (sqrt2-2)}=frac{sqrt2cdot (sqrt2)^7cdot (1-(sqrt2)^7)}{sqrt2-2}=$

$=frac{2cdot (sqrt2)^6cdot (1-2^3sqrt2)}{sqrt2(1-sqrt2)}=frac{8sqrt2cdot (1-8sqrt2)}{1-sqrt2}=frac{8sqrt2cdot (1-8sqrt2)(1+sqrt2)}{(1-sqrt2)(1+sqrt2)}=$

$=frac{8sqrt2cdot (1+sqrt2-8sqrt2-16)}{1-2}=-8sqrt2cdot (-15-7sqrt2)=8sqrt2cdot (15+7sqrt2)=\\=120sqrt2+112$