Question

Bn- прогрессия. b1+b2 / b2+b3 = 3, s3= 26. найти s6 ! вас !

Answer 1

Основные формулы:

$bn = b1*q^{n-1}$ 

$Sn = b1\frac{1-q^n}{1-q}$ 

$\frac{b1+b2}{b2+b3} = \frac{b1+b1q}{b1q+b1q^2} = \frac{b1(1+q)}{b1q(1+q)} = \frac{1}{q}= 3$ тогда q = 1/3

$S3 = b1\frac{1-q^3}{1-q} = b1\frac{(1-q)(1+q+q^2)}{1-q} = b1(1+q+q^2) = b1(1+\frac{1}{3}+\frac{1}{3^2}) = 26$ 

$\frac{13}{9}b1 = 26$ 

b1 = 18

$S6 = b1\frac{1-q^6}{1-q} = b1\frac{(1-q^2)(1+q^2+q^4)}{1-q} = b1\frac{(1-q)(1+q)(1+q^2+q^4)}{1-q} = b1(1+q)(1+q^2+q^4)$ 

Подставим найденные b1 и q в формулу:

$S6 = 18(1+\frac{1}{3})(1+\frac{1}{3^2}+\frac{1}{3^4}) = 18*\frac{4}{3}*\frac{91}{81} = \frac{6552}{243} = \frac{728}{27}$