1/1^2.3^2+3/5^2.7^2+...+12/23^2.25^2
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6 tháng 11 2016
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TL
0
30 tháng 10 2018
\(H=\frac{8}{1^2\cdot3^2}+\frac{16}{3^2\cdot5^2}+...+\frac{48}{11^2\cdot13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}\)
\(H=\frac{168}{169}\)
30 tháng 10 2018
Phương thiếu bước nhé
\(H=\frac{8}{1^2.3^2}+\frac{16}{3^2.5^2}+\frac{24}{5^2.7^2}+...+\frac{48}{11^2.13^2}\)
\(H=\frac{3^2-1^2}{1^2.3^2}+\frac{5^2-3^2}{3^2.5^2}+\frac{7^2-5^2}{5^2.7^2}+...+\frac{13^2-11^2}{11^2.13^2}\)
\(H=\frac{3^2}{1^2.3^2}-\frac{1^2}{1^2.3^2}+\frac{5^2}{3^2.5^2}-\frac{3^2}{3^2.5^2}+\frac{7^2}{5^2.7^2}-\frac{5^2}{5^2.7^2}+...+\frac{13^2}{11^2.13^2}-\frac{11^2}{11^2.13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{5^2}-\frac{1}{7^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}=1-\frac{1}{169}=\frac{168}{169}\)
Chúc bạn học tốt ~
20 tháng 7 2018
1. 25 : 5,7 = 250/57
2. 30:2.8.4 = 480
3. 20:2^2.14= 70
4. 125:5^3.170= 170
5. 64:2^5.30.4=240
6. (25:5^2.30): 15.7=14
bạn à! Nhiều quá mình ko làm hết được. sorry nha.^-^
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2019
\(\sqrt{1+\frac{8n^2-1}{\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{1+\frac{8n^2-1}{\left(4n^2-1\right)^2}}=\sqrt{\frac{\left(4n^2-1\right)^2+8n^2-1}{\left(4n^2-1\right)^2}}\)
\(=\sqrt{\frac{16n^4-8n^2+1+8n^2-1}{\left(4n^2-1\right)^2}}=\frac{4n^2}{4n^2-1}=1+\frac{1}{4n^2-1}=1+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(\Rightarrow S=1009+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=1009+\frac{1}{2}\left(1-\frac{1}{2019}\right)=...\)