Question

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

Answer 1

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

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

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