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Ta có : \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{31}{15^2.16^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+...+\dfrac{16^2-15^2}{15^2.16^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{15^2}-\dfrac{1}{16^2}\)
= \(1-\dfrac{1}{16^2}< 1\)
Xét số bất kì a. Ta sẽ chứng mỉnh (a + 1)2 - a2 = 2a + 1.
Thật vậy, ta có (a + 1)2 - a2 = a(a + 1) + (a + 1) - a2 = (a2 + a) + (a + 1) = 2a + 1 (đpcm).
Áp dụng ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\frac{1}{1^2}-\frac{1}{10^2}< 1\left(đpcm\right)\)
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
= \(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
= \(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
= \(1-\frac{1}{10^2}\)< 1
Vậy
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\) <1
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{2^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
\(\RightarrowĐPCM\)
Đặt \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(\Rightarrow A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{10^2}\)
\(\Rightarrow A=1-\frac{1}{10^2}\)
Mà \(1-\frac{1}{10^2}< 1.\)
\(\Rightarrow A< 1\left(đpcm\right).\)
Chúc bạn học tốt!
3/1^2.2^2 = 1/1^2 - 1/2^2
5/2^2.3^2 = 1/2^2 - 1/3^2
..............
19/9^2.10^2 = 1/9^2 - 1/10^2
=>3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 19/9^2.10^2 = 1/1^2 - 1/10^2
= 1/2 - 1/100 < 1
ko chắc lắm, bạn tham khảo nhé
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+..+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}<1\left(đpcm\right)\)
nhớ tick nhé
Đặt A là biểu thức trên
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{31}{15^2.16^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{31}{225.256}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{225}-\frac{1}{256}\)
\(=1-\frac{1}{256}=\frac{255}{256}< 1\)
Vậy...