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\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
<=> \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}=\frac{56}{673}\)
<=> \(4.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}\right)=4.\frac{56}{673}\)
<=> \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{n+4-3}{3.\left(n+4\right)}=\frac{224}{673}\Leftrightarrow\frac{n}{3.\left(n+4\right)}=\frac{224}{673}\)
<=> 673n = 224.3(n+4)
<=> 673n = 224.3.n + 224.3.4
<=> 673n = 672n + 2688
<=> 673n - 672n = 2688
<=> n = 2688
\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)
\(\Rightarrow4A=\)
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)
=> n + 4 = 2019
n = 2019 - 4
n = 2015
Ta có 1n2+4n=14(1n−1n+4)1n2+4n=14(1n−1n+4) Khi đó pt tương đương: 14(13−17+17−111+...+1n−1n+4)=5667314(13−17+17−111+...+1n−1n+4)=56673 ⟺13−1n+4=224673=>n=2015
A = \(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+\frac{1}{285}+\frac{1}{437}\)
A = \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{4}.\left(\frac{1}{11}-\frac{1}{15}\right)+\frac{1}{4}.\left(\frac{1}{15}-\frac{1}{19}\right)+\frac{1}{4}.\left(\frac{1}{19}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{23}\right)\)
A = \(\frac{1}{4}.\frac{20}{69}\)
A = \(\frac{5}{69}\)
Có: \(\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{n\left(n+4\right)}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{n\left(n+4\right)}=\dfrac{4.56}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{n}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{n+4}=\dfrac{1}{2019}\)
\(\Leftrightarrow n=2015\)
