tìm n thuộc N* biết :
1/21+1/77+1/165+...+1/n mũ 2+4n = 56/673
mấy bạn giúp mình với please
Tìm n thuộc N*, biết rằng 1/21 + 1/77 + 1/165 + ... + 1/n^2+4n = 56/673
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
- Tìm n thuộc N* biết rằng: 1/21 + 1/77 + 1/165 +...+ 1/n^2+4n = 56/673
tìm n thuộc N* biết rằng 1/21+1/77+1/165+...+1/n^2+4n=56/673
Tìm x thuộc N*, biết rằng
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
cái này là toán mà bạn, đâu phải vật lý
Tìm n\(\in\) N*, biết rằng:
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+.....+\frac{1}{n^2+4n}=\frac{56}{673}\)
nếu giải thích chi tiết mình cho 2 tick
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
<=> \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}=\frac{56}{673}\)
<=> \(4.\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n.\left(n+4\right)}\right)=4.\frac{56}{673}\)
<=> \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
<=> \(\frac{n+4-3}{3.\left(n+4\right)}=\frac{224}{673}\Leftrightarrow\frac{n}{3.\left(n+4\right)}=\frac{224}{673}\)
<=> 673n = 224.3(n+4)
<=> 673n = 224.3.n + 224.3.4
<=> 673n = 672n + 2688
<=> 673n - 672n = 2688
<=> n = 2688
Tìm n \(\in\) N* biết \(\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}\)
Có: \(\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+...+\dfrac{1}{n^2+4n}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{n\left(n+4\right)}=\dfrac{56}{673}\)
\(\Leftrightarrow\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{n\left(n+4\right)}=\dfrac{4.56}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{n}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{n+4}=\dfrac{224}{673}\)
\(\Leftrightarrow\dfrac{1}{n+4}=\dfrac{1}{2019}\)
\(\Leftrightarrow n=2015\)
Tìm n thuộc N* biết rằng:\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+1n}=\frac{56}{673}\)
Tìm n E N* biết:
\(\frac{1}{21}\)+\(\frac{1}{77}\)+\(\frac{1}{165}\)+...+\(\frac{1}{n^2+4n}\)=\(\frac{56}{673}\)
\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)
\(\Rightarrow4A=\)
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)
=> n + 4 = 2019
n = 2019 - 4
n = 2015
Giúp mình giải bài toán này với. Mình cám ơn.
Tính nhanh: B = 5/21 + 5/77 + 5/165 + ...+ ( 5/ (4n-1)(4n+3)
\(B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{1}{5}B=\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(B-\frac{1}{5}B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}-\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\)\(\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+\frac{4}{11}-\frac{4}{15}+...+\frac{4}{4n-1}-\frac{4}{4n+3}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{4n-3}\)
\(\frac{4}{5}B=\frac{16n-24}{12n-9}\)
\(B=\frac{\frac{16n-24}{12n-9}}{\frac{4}{5}}\)
\(B=\frac{20n-30}{12n-9}\)
B = \(\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}+\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}.\left(\frac{1}{3}-\frac{1}{4n+3}\right)=\frac{5}{12}-\frac{5}{4\left(4n+3\right)}=\frac{5}{12}-\frac{5}{16n+12}\)
sửa lại
\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)
\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{4n+3}\)
\(\frac{4}{5}B=\frac{4n}{12n+9}\)
\(B=\frac{\frac{4n}{12n+9}}{\frac{4}{5}}\)
\(B=\frac{5n}{12n+9}\)