Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
CMR A < 1
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14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 5 2023
Hôm nay olm sẽ hướng dẫn các em mẹo giải các bài toán dạng này như sau:
Ta thấy vế phải là \(\dfrac{1}{2}\) thì vế trái sẽ ≤ \(\dfrac{1}{2}\) - a ( a > 0)
Em biến đổi mẫu số các phân số lần lượt thành lũy thừa của các số tự nhiên liên tiếp. Sau đó rút gọn tổng các phân số đó thì sẽ chứng minh được em nhé.
A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)
A = \(\dfrac{1}{\left(1.2\right)^2}\)+\(\dfrac{1}{\left(2.2\right)^2}\)+\(\dfrac{1}{\left(2.3\right)^2}\)+...+\(\dfrac{1}{\left(2.50\right)^2}\)
A = \(\dfrac{1}{1^2.2^2}\)+\(\dfrac{1}{2^2.2^2}\)+\(\dfrac{1}{2^2.3^2}\)+...+\(\dfrac{1}{2^2.50^2}\)
A = \(\dfrac{1}{2^2}\)\(\times\)(\(\dfrac{1}{1^2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{50^2}\))
A = \(\dfrac{1}{4}\) \(\times\)(1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+...+\(\dfrac{1}{50.50}\))
Vì \(\dfrac{1}{1}\)> \(\dfrac{1}{2}\)>\(\dfrac{1}{3}\)>\(\dfrac{1}{4}\)>...>\(\dfrac{1}{50}\)
⇒ \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{50.50}\)<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...\(\dfrac{1}{49.50}\)
A = \(\dfrac{1}{4}\).(1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+\(\dfrac{1}{4.4}\)+..+\(\dfrac{1}{50.50}\)) < \(\dfrac{1}{4}\) .(1+\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+..+\(\dfrac{1}{49.50}\))
A < \(\dfrac{1}{4}\).(1+\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\))
A<\(\dfrac{1}{4}\).(2 - \(\dfrac{1}{50}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{200}\) < \(\dfrac{1}{2}\)
Vậy A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\) < \(\dfrac{1}{2}\) ( đpcm)
28 tháng 4 2017
a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)
\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)
49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)
28 tháng 4 2017
b) Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)
\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)
\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))
\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)
Lại có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
Mà :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)
\(\Rightarrow A>48\) \(\rightarrowđpcm\)
28 tháng 3 2017
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)
\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)
NA
Nguyễn An Ninh
VIP
25 tháng 4 2023
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
NA
Nguyễn An Ninh
VIP
25 tháng 4 2023
Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
Ta có : \(\dfrac{1}{2^2}=\dfrac{1}{2\times2}< \dfrac{1}{1\times2}\\ \dfrac{1}{3^2}=\dfrac{1}{3\times3}< \dfrac{1}{2\times3}\\ \dfrac{1}{4^2}=\dfrac{1}{4\times4}< \dfrac{1}{3\times4}\\ ...\\ \dfrac{1}{100^2}=\dfrac{1}{100\times100}< \dfrac{1}{99\times100}\)
\(\Rightarrow\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)
hay \(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{100}{100}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Vậy \(A< 1\)(đpcm)
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Vậy A<1