\(CMR:\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 2\)
Những câu hỏi liên quan
CMR: \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< \frac{173}{100}\)
Ta có: \(55+5\)1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 +.....+ 1/50^2 = 1/1^2 + 1/2^2 + (1/3^2 + 1/4^2 +....+ 1/50^2 )
< 1 + 1/4 + (1/2*3 + 1/3*4 +...+1/49*50) = 1 + 1/4 + (1/2 - 1/3 + 1/3 - 1/4+...+1/49 - 1/50 )
= 1,73 = 173/100 (dpcm)
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CMR \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{4}{9},A>\frac{1}{4}\)
A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\) CMR A < 2
AI ĐÚNG TK
Ta có : \(\frac{1}{1^2}=1\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
Vậy \(A< 2\)
CMR:
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2019\times2020}< 1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{3}{4}\)
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 2\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)
\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}...+\frac{1}{100!}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Leftrightarrow C< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow C< 2-\frac{1}{100}\)
\(\Leftrightarrow C< 2\left(đpcm\right)\)
Cmr:\(50< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100-1}}< 100\)
Gọi biểu thức trên là A.
Chứng minh A > 50
\(A=1+\frac{1}{2}+\left(\frac{1}{2^1+1}+\frac{1}{2^2}\right)+\left(\frac{1}{2^2+1}+\frac{1}{6}+...+\frac{1}{2^3}\right)+...+\left(\frac{1}{^{2^{100-2}+1}}+...+\frac{1}{2^{100-1}}\right)\\ \)
\(A>1+\frac{1}{2}+\frac{1}{2^2}.2+\frac{1}{2^3}.2^2+...+\frac{1}{2^{100-1}}2^{100-2}\)
\(A>\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\)
\(< =>A>\frac{100}{2}=50\)
Chứng minh A<100
\(A=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2^2}+\frac{1}{5}+...+\frac{1}{7}\right)+....+\left(\frac{1}{2^{100-2}}+\frac{1}{2^{100-2}+1}+...+\frac{1}{2^{100-1}-1}\right)\)-\(\frac{1}{2^{100-1}}\)
\(A< 1+\frac{1}{2}.2+\frac{1}{2^2}.2^2+...+\frac{1}{2^{100-2}}.2^{100-2}+\frac{1}{2^{100-1}}\)
\(A< 1+1+1+...+1+\frac{1}{2^{100-1}}\)
\(A< 1.99+\frac{1}{2^{100-1}}< 99+1=100\)
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ta có : 1+1/2+1/3+....+1/2^100-1
= 1/2x2 +1/3x2 +1/4x2 +...+ 1/2^100 x2
= 2x(1/2+1/3+1/4+...+1/2^100)
=.................... làm đến đây mk tịt
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Xem thêm câu trả lời
frac{1}{2^2}+frac{1}{4^2}+frac{1}{6^2}+..+frac{1}{100^2}frac{1}{2^2}left(1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2}right)Có frac{1}{2^2} frac{1}{1.2}1-frac{1}{2} frac{1}{3^2} frac{1}{2.3}frac{1}{2}-frac{1}{3}....v........v............ frac{1}{50^2} frac{1}{49.50}frac{1}{49}-frac{1}{50}Cộng lại 1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2} 1+1-frac{1}{2}+frac{1}{2}-frac{1}{3}+...+frac{1}{49}-frac{1}{50}2-frac{1}{50}Rightarrow VT frac{1}{2^2}left(2-frac{1}{50}right)frac{1}{2...
Đọc tiếp
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\) \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)....v........v............ \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
Cộng lại \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}\)
\(\Rightarrow VT< \frac{1}{2^2}\left(2-\frac{1}{50}\right)=\frac{1}{2}-\frac{1}{2^2.50}< \frac{1}{2}\left(Đpcm\right)\)
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ủa toán lớp mấy chứ ko phải lớp 1
uk ko phải toán lớp 1
\(S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
CMR: S < 2
P/s: Ko tiếp loại Spam
Ta có: \(S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(\Rightarrow S< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow S< 2-\frac{1}{50}\)
Vậy S < 2
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Bài 1 : Cho A = \(\frac{1}{^{1^2}}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\) . CMR : A < 2
Bài 2 : Cho B = \(2^1+2^2+2^3+2^4+...+2^{30}\). CMR : B chia hết cho 21
Bai 2 :
Ta co :
B = [ 2^1 + 2^2 + 2^3 + 2^4 + 2^5 = 2^6 ] + .... + [ 2^25 + 2^26 + 2^27 + 2^28 +2^29 +2^30 ]
= 2[1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 ] +.....+ 2^25[ 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 ]
= 2 . 63 +.... + 2^25 . 63
= 63 [2 + ..... + 2^25 ] chia het cho 21
Vay B chia het cho 21
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Bai 1 :
Ta co :
A = 1/1 + 1/2^2 + 1/3^3 + 1/4^4 + .... + 1?50^2 < 1/1 + 1/1.2 + 1/2.3 + ..... + 1/49.50
=>1 + 1/1 - 1/2 +1/2 -1/3 + .... +1/449 - 1/50
=> 1 + 1/1 - 1/50
=> 1 + 49/50
=> 99/50 < 2
Vay 1 < 2
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bai 1 minh ket luan nham
A < 2
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A frac{3}{4}+frac{8}{9}+frac{15}{16}+....+frac{2499}{2500}A1-frac{1}{4}+1-frac{1}{9}+1-frac{1}{16}+....+1-frac{1}{2500}Aleft(1+1+1+.....+1right)-left(frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{50^2}right)A49-left(frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{50^2}right)do frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{50^2}0 nên 490bài trên iu cầu CMR A 49 thì mk lm đúng chưa ạ. Đây là đề thi quận mk đó ạ
Đọc tiếp
A= \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}\)
A=\(1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+....+1-\frac{1}{2500}\)
A=\(\left(1+1+1+.....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
A=\(49-\)\(\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
do \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)>0 nên 49<0
bài trên iu cầu CMR A < 49 thì mk lm đúng chưa ạ. Đây là đề thi quận mk đó ạ
(: ko bít. tui giỏi tiếng anh nhưng ngu toán lắm
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