1^2/1.3+2^2/1.3+................+99^2/197.199
Những câu hỏi liên quan
Tính M=1^2/1.3+2^2/3.5+3^2/5.7+...+99^2/197.199
Tính \(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+........+\frac{99^2}{197.199}\)
Tính M=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{99^2}{197.199}\)
M=1^2/1.3+2^2/3.5+3^2/5.7+.....+99^2/197.199
Các bạn tìm M giúp mik nhé mik cần gấp bạn nào giải đúng và nhanh nhất mik tik cho
Hướng dẫn:
\(M=\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{99^2}{197.199}\)
\(\Rightarrow4M=\frac{1.4}{1.3}+\frac{4.4}{3.5}+\frac{9.4}{5.7}+...+\frac{9801.4}{197.199}\)
\(\Rightarrow4M=\frac{2.2}{1.3}+\frac{4.4}{3.5}+\frac{6.6}{5.7}+...+\frac{198.198}{197.199}\)
Đến đoạn này bạn đưa về dạng tổng quát nhé:
\(\frac{n^2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{4}+\frac{1}{8\left(2n-1\right)}-\frac{1}{8\left(2n+1\right)}\) (Tự phân tích)
Sau đó thay vào A. Kết quả tìm được là \(A=\frac{1}{8}-\frac{1}{8.2013}+\frac{1006}{4}=251,6249379\)
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11 tháng 7 2021 lúc 16:23
tính:a)1/2022-5/2.4-5/4.6-5/6.8.....5/2020.2022
b)2^2/1.3+2^2/3.5+...+2^2/197.199
a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)
\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)
\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)
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b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
\(=2\left(1-\dfrac{1}{199}\right)\)
\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)
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\(B=\frac{2^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2^2}{195.197}+\frac{2^2}{197.199}\)
Mọi người giúp mình làm nhé, mình đang cần bài này gấp
\(B=\frac{2^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2^2}{195.197}=2\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{195.197}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{195}-\frac{1}{197}\right)=2\left(1-\frac{1}{197}\right)=2.\frac{196}{197}=\frac{392}{197}\)
1/1.3+1/3.5+1/5.7+...+1/(2.x-1)(2.x+1)=49/99
=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99
=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99
=>1-1/(2x+1)=98/99
=>1/(2x+1)=1/99
=>2x+1=99
=>x=49
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\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+.....+\left|x+\dfrac{1}{197.199}\right|=100x\)
chứng minh rằng 1/2+1.3+1/4+...+1/100 > 99/100
