Question

а) b1=0,(3)
$q = \sqrt{3}$
S6-?

б)
$b1 = {7}^{ - 1}$
$q = \sqrt{7}$
S6-?​

Answer 1

а)  

$b_1=0,(3)=\frac{3}{9} =\frac{1}{3}$

$q=\sqrt{3}$

$S_6-?$

Решение

$S_n=\frac{b_1(1-q^n}{1-q}$

$S_6=\frac{\frac{1}{3}*(1-(\sqrt{3})^6)}{1-\sqrt{3}}=\frac{1-(\sqrt{3})^6}{3*(1-\sqrt{3})}=\frac{1-3^3}{3*(1-\sqrt{3})}=\frac{1-27}{3*(1-\sqrt{3})}=\frac{-26}{3*(1-\sqrt{3})}=$

  $=\frac{-26*(1+\sqrt{3} )}{3*(1-\sqrt{3})*(1+\sqrt{3} )}=\frac{-26*(1+\sqrt{3} )}{3*(1^2-(\sqrt{3})^2)}= \frac{-26*(1+\sqrt{3} )}{3*(1-3)}= \frac{-26*(1+\sqrt{3} )}{3*(-2)}= \frac{13(1+\sqrt{3} )}{3}$

Ответ:    $\frac{13(1+\sqrt{3} )}{3}$

б)

$b_1=7^{-1}=\frac{1}{7}$

$q=\sqrt{7}$

$S_6-?$

Решение

$S_n=\frac{b_1(1-q^n}{1-q}$

$S_6=\frac{\frac{1}{7}*(1-(\sqrt{7})^6}{1-\sqrt{7} }=\frac{1*(1-(\sqrt{7})^6}{7*(1-\sqrt{7}) }=\frac{1-7^3}{7*(1-\sqrt{7}) }=\frac{1-343}{7*(1-\sqrt{7}) }=\frac{-342*(1+\sqrt{7} )}{7*(1-\sqrt{7})(1+\sqrt{7} ) }=$

$=\frac{-342*(1+\sqrt{7} )}{7*(1^2-(\sqrt{7})^2) }=\frac{-342*(1+\sqrt{7} )}{7*(1-7) }=\frac{-342*(1+\sqrt{7} )}{7*(-6) }=\frac{57*(1+\sqrt{7} )}{7}$

Ответ:    $\frac{57*(1+\sqrt{7} )}{7}$