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a, \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow1< 1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
Mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2\Rightarrow A< 2\left(đpcm\right)\)
b, B = 2 + 22 + 23 +...+ 230
= (2+22+23+24+25+26)+...+(225+226+227+228+229+230)
= 2(1+2+22+23+24+25)+...+225(1+2+22+23+24+25)
= 2.63+...+225.63
= 63(2+...+225)
Vì 63 chia hết cho 21 nên 63(2+...+225) chia hết cho 21
Vậy B chia hết cho 21
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(............\)
\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\)\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\)\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\)\(A< 1+1-\frac{1}{50}< 1+1\)
\(\Rightarrow\)\(A< 2\)
Vậy \(A< 2\)
Chúc bạn học tốt ~
CM A < 2
=> CM \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\)
Ta thấy: \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{2}=\frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{6}=\frac{1}{2.3}\)
Và cứ thế,....
\(\frac{1}{50^2}=\frac{1}{2500}< \frac{1}{2450}=\frac{1}{49.50}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\)
=> \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1^2}=1+1=2\)
=>ĐPCM
1/1^2 < 1/1x2 < 1 - 1/2
1/2^2 < 1/2x3 < 1/2 -1/3
...
1/50^2 < 1/50x51 < 1/50 - 1/51
Tính tổng ta có A <1-1/51 <2
Bài 1:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
Bài 2:
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vậy A < 2
Bài 3:
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Bài 4:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)
A=1-1/2+1/2-1/3+.............................1/49-1/50
A=1-1/50
A=49/50
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
CHO \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}.\)CHỨNG MINH A<2
\(\frac{1}{2^2}< \frac{1}{1.2}\)
...................\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)
1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50
ta có:
=> 1/1^2+1/2*3+1/3*4+...+1/49*50
<=> 1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
<=> 1-1/50 < 2
=> A < 2
Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\) A < \(1+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow\) A < \(1+\left(1-\frac{1}{50}\right)\)
\(\Rightarrow\) A < 1 + 49/50
Mà 1+49/50 < 2 nên A < 1+49/50 < 2
\(\Rightarrow\) A < 2
\(A=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}=B\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}< 1\)
Vì \(A< 1+B\)mà \(B< 1\)nên \(B+1< 2\)do đó \(A< 2\)
Vậy \(A< 2\)
1/12+1/22+....+1/502<1/1+1/1x2+1/2x3+....+1/49x50=1-1/50=49/50<2
=>A<2(đpcm)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)
\(=\frac{1}{1.1}+\frac{1}{2.2}+...+\frac{1}{50.50}\)
\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
Vậy A < 2
TỔNG A BAO GỒM CÁC SỐ HẠNG ĐC BIỂU DIỄN DƯỚI DẠNG PHÂN SỐ CÓ MẪU > TỬ NÊN TỔNG A CŨNG ĐC BIỂU DIỄN DƯỚI DẠNG CÁC PHÂN SỐ CÓ MẪU > TỬ HAY(TỬ KO CHIA HẾT CHO MẪU)KO THỂ VIẾT DƯỚI DẠNG SỐ TỰ NHIÊN.
VẬY: A<2.