Question

Сроочнооооо, спасайте буду очень блогодарнаааа​

Answer 1

Объяснение:

$1.\\a)\ \lim_{n \to \infty} (1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...+\frac{(-1)^{n-1}}{3^{n-1}})\\ 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...+\frac{(-1)^{n-1}}{3^{n-1}}\\q=\frac{b_2}{b_1}=\frac{-\frac{1}{3} }{1} =-\frac{1}{3} .\\b_1=1.\\S=\frac{b_1}{1-q}=\frac{1}{1-(-\frac{1}{3}) } =\frac{1}{1+\frac{1}{3} }=\frac{1}{1\frac{1}{3} }=\frac{1}{\frac{4}{3} }=\frac{3}{4} .\\ \lim_{n \to \infty} \frac{3}{4} =\frac{3}{4}.$

$b)\ \lim_{n \to \infty} (\frac{1}{n^2}+\frac{2}{n^2} +\frac{3}{n^2}+...+\frac{n-1}{n^2})\\\frac{1}{n^2}+\frac{2}{n^2} +\frac{3}{n^2}+...+\frac{n-1}{n^2}\\q=\frac{\frac{n-1}{n^2} }{\frac{n-2}{(n-1)^2} } =\frac{(n-1)*(n-1)^2}{n^2*(n-2)}=\frac{(n-1)^3}{n^2*(n-2)}.\\b_1=\frac{1}{n^2}.\\S=\frac{\frac{1}{n^2} }{1-\frac{(n-1)^3}{n^2*(n-2)} } =\frac{\frac{1}{n^2} }{\frac{n^3-2n^2-n^3+3n^2-3n+1}{n^2*(n-2)} }=\frac{n-2}{n^2-3n+1} .$

$\lim_{n \to \infty} \frac{n-2}{n^2-3n+1}=|\frac{:n^2}{:n^2}|= \lim_{n \to \infty} \frac{\frac{1}{n}-\frac{2}{n^2} }{1-\frac{3}{n}+\frac{1}{n^2} } =\frac{0-0}{1-0+0}=\frac{0}{1}=0.$

$2.\ \\a)\ a_{10}=21\ \ \ \ S_{19}=?\\S_{19}=\frac{2a_1+18d}{2} *19=\frac{2*(a_1+9d)}{2}*19=(a_1+9d)*19=a_{10}*19=21*19=399.$

Ответ: S₁₉=399.

$b)\ a;\ 2a+2;\ 3a+4;\ ....\ \ \ \ S_{10}=255\ \ \ \ a=?\\d=2a+2-a=a+2.\\S_{10}=\frac{2*a+9*(a+2)}{2}*10=255\\\frac{2a+9a+18}{2}*10=255\\(11a+18)*5=255\ |:5\\11a+18=51\\11a=33\ |:11\\a=3.$

Ответ: a=3.

$3.\\a)\ \left \{ {{b_1+b_2+b_3+b_4=15} \atop {b_5+b_6+b_7+b_8=240}} \right. \ \ \ \ \left \{ {{b_1+b_1q+b_1q^2+b_1q^3=15} \atop {b_1q^4+b_1q^5+b_1q^6+b_1q^7=240}} \right. \ \ \ \ \left \{ {{b_1*(1+q+q^2+q^3)=15} \atop {b_1q^4*(1+q+q^2+q^3)=240}} \right.$

Разделим второе уравнение на первое:

$q^4=\frac{240}{15}=16\\q^4=2^4\\q=б2. \\1) \ q=2.\\b_1*(1+2+2^2+2^3)=15\\b_1*(1+2+4+8)=15\\15*b_1=15\ |:15\\b_1=1.\\S_{6}=1*\frac{2^6-1}{2-1} =\frac{64-1}{1}=63.\\S_6=63.\\2)\ q=-2.\\b_1*(1-2+4-8)=15\\b_1*(-5)=15\\ -5*b_1=15\ |:(-5)\\b_1=-3.\\S_6'=-3*\frac{(-2)^6-1}{-2-1} =\frac{-3*63}{-3} =63.$

Ответ: S₆=63.

$b)\ \frac{\sqrt{2}+1 }{\sqrt{2}-1 } ;\ \frac{1}{2-\sqrt{2} } ;\ \frac{1}{2};\ ...\ \ \ \ S=?\\q=\frac{\frac{1}{2} }{\frac{1}{2-\sqrt{2} } }=\frac{2-\sqrt{2} }{2}=1-\frac{\sqrt{2} }{2} . \\S=\frac{\frac{\sqrt{2}+1 }{\sqrt{2} -1} }{1-1+\frac{\sqrt{2} }{2} } =\frac{\frac{\sqrt{2}+1 }{\sqrt{2}-1 } }{\frac{\sqrt{2} }{2} }=\frac{2*(\sqrt{2}+1) }{\sqrt{2}*(\sqrt{2}-1) } =\frac{2*(\sqrt{2}+1)^2 }{\sqrt{2} }=\\=\sqrt{2}*(2+2\sqrt{2}+1)=\sqrt{2}*(3+2\sqrt{2}) = 4+3\sqrt{2}.$

Ответ: S=4+3√2.