Câu hỏi của hunh lê

Question

Tìm n $\in$ N* biết $\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}$

Anh Triêt · Accepted Answer

Có: $\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}$

$\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{n\left(n+4\right)}=\dfrac{56}{673}$

$\Leftrightarrow\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{n\left(n+4\right)}=\dfrac{4.56}{673}$

$\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{n}-\dfrac{1}{n+4}=\dfrac{224}{673}$

$\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{n+4}=\dfrac{224}{673}$

$\Leftrightarrow\dfrac{1}{n+4}=\dfrac{1}{2019}$

$\Leftrightarrow n=2015$Câu trả lời: Có: $\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}$

$\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{n\left(n+4\right)}=\dfrac{56}{673}$

$\Leftrightarrow\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{n\left(n+4\right)}=\dfrac{4.56}{673}$

$\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{n}-\dfrac{1}{n+4}=\dfrac{224}{673}$

$\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{n+4}=\dfrac{224}{673}$

$\Leftrightarrow\dfrac{1}{n+4}=\dfrac{1}{2019}$

$\Leftrightarrow n=2015$