Question

1.Найдите знаменатель геометрической прогрессии:
а)3, 9, 27,
б)1/√2, 1/4, 1/8√2, ...

2.Найдите сумму первых пяти членов геометрической прогрессии (bₓ), у которой b₁=14, q=1/2

3. Найдите сумму x+1+1/x+...+1/x⁶

Answer 1

$1)\ \ a)\ \ \{b_{n}\}:\ \ 3\ ;\ 9\ ;\ 27\ ;\ ...\\\\q=\dfrac{b_{n+1}}{b_{n}}\ \ \to \ \ \ q=\dfrac{b_2}{b_1}=\dfrac{9}{3}=3\\\\Proverka:\ \ b_3=b_2q=9\cdot 3=27\\\\b)\ \ \{b_{n}\}:\ \ \dfrac{1}{\sqrt2}\ ;\ \dfrac{1}{4}\; ;\ \dfrac{1}{8\sqrt2}\ ;\ ...\\\\q=\dfrac{1/4}{1/\sqrt2}=\dfrac{\sqrt2}{4}=\dfrac{1}{2\sqrt2}\\\\q=\dfrac{1/8\sqrt2}{1/4}=\dfrac{4}{8\sqrt2}=\dfrac{1}{2\sqrt2}\\\\Proverka:\ \ b_3=b_2q=\dfrac{1}{4}\cdot \dfrac{1}{2\sqrt2}=\dfrac{1}{8\sqrt2}$

$2)\ \ b_1=14\ \ ,\ \ q=\dfrac{1}{2}\\\\S_5=\dfrac{b_1\, (q^5-1)}{q-1}=\dfrac{14\, (\frac{1}{32}-1)}{\frac{1}{2}-1}=\dfrac{-14\cdot \frac{31}{32}}{-\frac{1}{2}}=\dfrac{14\cdot 31\cdot 2}{32}=\dfrac{7\cdot 31}{8}=\\\\=\dfrac{217}{8}=27,125$

$3)\ \ x+1+\dfrac{1}{x}+...+\dfrac{1}{x^6}=x+1+\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+\dfrac{1}{x^4}+\dfrac{1}{x^5}+\dfrac{1}{x^6}\\\\q=\dfrac{b_2}{b_1}=\dfrac{1}{x}\\\\S_8=\dfrac{x\cdot (\dfrac{1}{x^8}-1)}{\dfrac{1}{x}-1}=\dfrac{x\cdot (1-x^8)\cdot x}{x^8\cdot (1-x)}=\dfrac{(1-x^8)}{x^6(1-x)}=\dfrac{(1-x^4)(1+x^4)}{x^6(1-x)}=\\\\\\=\dfrac{(1-x)(1+x)(1+x^2)(1+x^4)}{x^6(1-x)}=\dfrac{(1+x)(1+x^2)(1+x^4)}{x^6}$