cho A = \(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{50^2}\)
CMR A>\(\frac{1}{4}\)va A<\(\frac{4}{9}\)
CMR \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{4}{9},A>\frac{1}{4}\)
Cho A = 1 + \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2^{100}-1}\)
CMR: A > 50
Cho:
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2^{100}-1}\)
CMR: A>50
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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
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
bai 1 minh ket luan nham
A < 2
cho a,b,c >0 va abc=1.
CMR \(\frac{1}{ab+a+2}+\frac{1}{bc+c+2}+\frac{1}{ca+a+2}\le\frac{3}{4}\)
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 \(A< 2\)
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
Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{50}\)
CMR A ko phải là số tự nhiên