Question

Найти значение суммы

Answer 1

$1-\frac{1}{3} =\frac{3}{1\cdot 3} -\frac{1}{3}=\frac{2}{1\cdot 3}$      ⇒    $\frac{2}{1\cdot 3}=1-\frac{1}{3}$    ⇒

$\frac{1}{1\cdot 3}=(1-\frac{1}{3} )\cdot \frac{1}{2}$

$\frac{1}{3} -\frac{1}{5} =\frac{5}{3\cdot 5} -\frac{3}{3\cdot 5}=\frac{2}{3\cdot 5}$      ⇒    $\frac{2}{3\cdot 5}=\frac{1}{3} -\frac{1}{5}$    ⇒

$\frac{1}{3\cdot 5}=(\frac{1}{3} -\frac{1}{5})\cdot \frac{1}{2}$

$\frac{1}{2n-1} -\frac{1}{2n+1} =\frac{2n+1-2n+1}{(2n-1)\cdot (2n+1)} =\frac{2}{(2n-1)\cdot (2n+1)}$      ⇒    $\frac{2}{(2n-1)\cdot (2n+1)}=\frac{1}{2n-1} -\frac{1}{2n+1}$    ⇒

$\frac{1}{(2n-1)\cdot (2n+1}=(\frac{1}{2n-1} -\frac{1}{2n+1})\cdot \frac{1}{2}$

Получаем

$\frac{1}{1\cdot 3}+ \frac{1}{3\cdot 5}+...+\frac{1}{(2n-1)\cdot (2n+1)}=(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1})\cdot \frac{1}{2}=\\\\=(1-\frac{1}{2n+1})\cdot \frac{1}{2}= \frac{n}{2n+1}$

2)

Аналогично

$\frac{1}{1\cdot 4}+ \frac{1}{4\cdot 7}+...+\frac{1}{(3n-2)\cdot (3n+1)}=(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{3n-2}-\frac{1}{3n+1})\cdot \frac{1}{3}=\\\\=(1-\frac{1}{3n+1})\cdot \frac{1}{3}= \frac{n}{3n+1}$

Answer 2

Ответ:

Для нахождения коэффициентов при разложении правильной дроби на сумму простейших дробей пользуемся методом неопределённых коэффициентов .

$1)\ \ \displaystyle \frac{1}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}=\frac{A(2n+1)+B(2n-1)}{(2n-1)(2n+1)}\ \Rightarrow \\\\\\1=\underbrace{0\cdot n+1}=2An+A+2Bn+B=\underbrace{(2A+2B)\cdot n+(A-B)}_{0\cdot n+1}\ \ \Rightarrow \\\\2A+2B=0\ \ ,\ \ A=-B\ \\A-B=1\ \ ,\ \ \ \ \ A=1+B\ ,\ \ -B=1+B\ ,\ \ 2B=-1\ ,\ \ B=-\dfrac{1}{2}\ ,\ \ A=\dfrac{1}{2}\\\\\frac{1}{(2n-1)(2n+1)}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}=\frac{1}{2}\cdot \Big(\frac{1}{2n-1}-\frac{1}{2n+1}\Big)\ ;$

$\displaystyle \star\ \ \frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+...+\frac{1}{(2n-3)(2n-1)}+\frac{1}{(2n-1)(2n+1)}=\\\\\\=\frac{1}{2}\cdot \Big(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-3}-\frac{1}{2n-1}+\frac{1}{2n-1}+\frac{1}{2n+1}\Big)=\\\\\\=\frac{1}{2}\cdot \Big(\frac{1}{1}-\frac{1}{2n+1}\Big)=\frac{1}{2}\cdot \frac{2n}{2n+1}=\boxed{\ \frac{n}{2n+1}\ }$

$2)\ \ \displaystyle \frac{1}{(3n-2)(3n+1)}=\frac{A}{3n-2}+\frac{B}{3n+1}=\frac{A(3n+1)+B(3n-2)}{(3n-2)(3n+1)}\ \Rightarrow \\\\\\1=\underbrace{0\cdot n+1}=3An+A+3Bn-2B=\underbrace{(3A+3B)\cdot n+(A-2B)}_{0\cdot n+1}\ \ \Rightarrow \\\\3A+3B=0\ \ ,\ \ A=-B\ \\A-2B=1\ \ ,\ \ \ \ \ A=1+2B\ ,\ \ -B=1+2B\ ,\ \ 3B=-1\ ,\ \ B=-\dfrac{1}{3}\ ,\ \ A=\dfrac{1}{3}\\\\\frac{1}{(3n-2)(3n+1)}=\frac{1}{3(3n-2)}-\frac{1}{3(3n+1)}=\frac{1}{3}\cdot \Big(\frac{1}{3n-2}-\frac{1}{3n+1}\Big)\ ;$

$\displaystyle \star \ \ \frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+...+\frac{1}{(3n-5)(3n-2)}+\frac{1}{(3n-2)(3n+1)}=\\\\\\=\frac{1}{3}\cdot \Big(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\frac{1}{3n-5}-\frac{1}{3n-2}+\frac{1}{3n-2}+\frac{1}{3n+1}\Big)=\\\\\\=\frac{1}{3}\cdot \Big(\frac{1}{1}-\frac{1}{3n+1}\Big)=\frac{1}{3}\cdot \frac{3n}{2n+1}=\boxed{\ \frac{n}{3n+1}\ }$