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Câu a) Mik chữa lại một chút
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\);.......; \(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Suy ra: \(VT< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy : \(VT+1< 1+1=2\)
Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)
ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
hk tốt #
Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
\(\Rightarrow\)Bài toán được chứng minh
Ta có: \(\frac{1}{1^2}=1\)
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
...
\(\frac{1}{50^2}<\frac{1}{49.50}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=> A < 1 - 1/50 = 49/50
Mà 49/50 < 50/50 = 1 < 2
=> A < 2 (Đpcm).
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
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Yêu cầu của bài là gì vậy. Tính A? hay Chứng minh A < 2 hoặc chứng minh A không phải là số nguyên
Chứng minh A < 2
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
Vậy A < 2
B < 1+1+1/2.3+1/3.4+...+1/62.63
B < 2+(1/2-1/3+1/3-1/4+...+1/62-1/63)
B < 2+(1/2-1/63)
B < 2+61/126 suy ra B < 2 và 6/126
Mà 2 + 61/126 <6
Suy ra B< 2+6/126<6 suy tiếp B < 6
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
S = ( 1 - \(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))....(1-\(\dfrac{1}{50^2}\))
S = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{50^2-1}{50^2}\)
Vì em lớp 6 nên phải làm thêm bước này nữa:
Ta có
n2 - 1 = n2 - n + n - 1 = (n2 - n) + (n - 1) = n(n-1) + (n-1) =(n-1)(n+1)
Áp dụng công thức vừa chứng minh trên vào tổng S ta có:
S = \(\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}\).\(\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}\)....\(\dfrac{\left(50-1\right)\left(50+1\right)}{50^2}\)
S = \(\dfrac{1.3}{2^2}\).\(\dfrac{2.4}{3^2}\)......\(\dfrac{49.51}{50^2}\)
S = \(\dfrac{\left(3.4.5.6....49\right)^2.1.2.50.51}{\left(3.4.5.6...49\right)^2.2.2.50.50}\)
S = \(\dfrac{1}{2}\) . \(\dfrac{51}{50}\)
S = \(\dfrac{51}{100}\)