Question

Вычислите
$\displaystyle \bigg ( \frac{1^2}{1\cdot 3} + \frac{2^2}{3\cdot 5 } + \frac{3^2}{5\cdot 7 } + ... + \frac{10^2}{19\cdot 21} \bigg)\cdot \bigg ( 2 - \frac{1}{11} \bigg )$

Answer 1

Ответ:

5.

Объяснение:

Воспользуемся формулой (которая проверяется непосредственно):

                                  $\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right).$

Имеем:                 $\dfrac{1^2}{1\cdot 3}+\dfrac{2^2}{3\cdot 5}+\dfrac{3^2}{5\cdot 7}+\ldots +\dfrac{10^2}{19\cdot 20}=$

               $=\dfrac{1}{2}\left(\dfrac{1^2}{1}-\dfrac{1^2}{3}+\dfrac{2^2}{3}-\dfrac{2^2}{5}+\dfrac{3^2}{5}-\dfrac{3^2}{7}+\ldots +\dfrac{10^2}{19}-\dfrac{10^2}{21}\right)=$

$=\dfrac{1}{2}\left(1+\dfrac{2^2-1^2}{3}+\dfrac{3^2-2^2}{5}+\dfrac{4^2-3^2}{7}+\ldots +\dfrac{10^2-9^2}{19}-\dfrac{100}{21}\right)=$

$=\dfrac{1}{2}\left(1+\dfrac{(2-1)(2+1)}{3}+\dfrac{(3-2)(3+2)}{5}+\ldots +\dfrac{(10-9)(10+9)}{19}-\dfrac{100}{21}\right)=$

                              $=\dfrac{1}{2}\left(1+1+1+1+\ldots +1-\dfrac{100}{21}\right)=$

                      $=\dfrac{1}{2}\left(10-\dfrac{100}{21}\right)=\dfrac{1}{2}\cdot \dfrac{210-100}{21}=\dfrac{110}{2\cdot 21}=\dfrac{11\cdot 5}{21}.$

 

 

Answer 2

Ответ:

$5$

Объяснение:

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

$\displaystyle \bigg ( \frac{1^2}{1\cdot 3} + \frac{2^2}{3\cdot 5 } + \frac{3^2}{5\cdot 7 } + ... + \frac{10^2}{19\cdot 21} \bigg)\cdot \bigg ( 2 - \frac{1}{11} \bigg )=$

$\displaystyle \bigg (1^2\cdot \left( \frac{1}{2}-\frac{1}{6}\right) +2^2\cdot \left( \frac{1}{6}-\frac{1}{10}\right) +3^2\cdot \left(\frac{1}{10}-\frac{1}{14} \right) + ... + 10^2\cdot \left(\frac{1}{38}-\frac{1}{42} \right) \bigg)\cdot \bigg ( 2 - \frac{1}{11} \bigg )=$

$\displaystyle \bigg ( \frac{1}{2}-\frac{1}{6} + \frac{4}{6}-\frac{4}{10}+\frac{9}{10}-\frac{9}{14} + ... + \frac{100}{38}-\frac{100}{42} \bigg)\cdot \bigg ( 2 - \frac{1}{11} \bigg )=$

$\displaystyle \bigg ( \frac{1}{2} + \frac{3}{6}+\frac{5}{10} + ... -\frac{100}{42} \bigg)\cdot \bigg ( 2 - \frac{1}{11} \bigg )=$

$\displaystyle \bigg ( \frac{1}{2} + \frac{1}{2}+\frac{1}{2} + ... -\frac{50}{21} \bigg)\cdot \frac{21}{11}=$

$\displaystyle \bigg (10\cdot \frac{1}{2}-\frac{50}{21} \bigg)\cdot \frac{21}{11}=$

$\displaystyle \bigg (5-\frac{50}{21} \bigg)\cdot \frac{21}{11}=$

$\displaystyle \bigg (\frac{105}{21}-\frac{50}{21} \bigg)\cdot \frac{21}{11}=5$

$\displaystyle \frac{55}{21}\cdot \frac{21}{11}=5$