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NV
1
23 tháng 5 2019
\(A=\frac{7}{4}+\frac{17}{9}+\frac{31}{16}+....+\frac{4999}{2500}\)
\(A=\frac{3+4}{2^2}+\frac{8+9}{3^2}+\frac{15+16}{4^2}+....+\frac{2499+2500}{50^2}\)
\(A=\frac{\left(4-1\right)+4}{2^2}+\frac{\left(9-1\right)+9}{3^2}+\frac{\left(16-1\right)+16}{4^2}+...+\frac{\left(2500-1\right)+2500}{50^2}\)
\(A=\frac{4+4-1}{2^2} +\frac{9+9-1}{3^2}+\frac{16+16-1}{4^2}+...+\frac{2500+2500-1}{50^2}\)
\(A=\frac{2^2+2^2-1}{2^2}+\frac{3^2+3^2-1}{3^2}+\frac{4^2+4^2-1}{4^2}+...+\frac{50^2+50^2-1}{50^2}\)
\(A=\frac{2\times2^2-1}{2^2}+\frac{2\times3^2-1}{3^2}+\frac{2\times4^2-1}{4^2}+...+\frac{2\times50^2-1}{50^2}\)
\(A=\left(\frac{2\times2^2}{2^2}-\frac{1}{2^2}\right)+\left(\frac{2\times3^2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{2\times4^2}{4^2}-\frac{1}{4^2}\right)+...+\left(\frac{2\times50^2}{50^2}-\frac{1}{50^2}\right)\)
\(A=\left(\frac{2\times2^2}{2^2}+\frac{2\times3^2}{3^2}+\frac{2\times4^2}{4^2}+...+\frac{2\times50^2}{50^2}\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
\(A=\left(2+2+2+...+2\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
49 số 2
\(A=98-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
\(>98-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{49\times50}\right)\)
\(=98-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=98-\left(1-\frac{1}{50}\right)\)
\(=98-1+\frac{1}{50}\)
\(=97+\frac{1}{50}>97\)
\(\Rightarrow A>97\left(ĐPCM\right)\)
Nhớ tích cho mk nha thank for watching
L
1
13 tháng 5 2017
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
\(=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow B=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1=48\)
\(\Rightarrow\) B > 48 (đpcm)
6 tháng 5 2017
a) gọi \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
gọi \(B=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(\Rightarrow A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)
b) Ta thấy \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\), \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)
Do đó : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3=\frac{1}{2}\)
c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Ta thấy vế trong ngoặc nhỏ hơn 1
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>48\)
KT
1
LB
4 tháng 5 2018
\(H=2+\dfrac{4-1}{4}+\dfrac{9-1}{9}+\dfrac{16-1}{16}+..+\dfrac{2500-1}{2500}\)\(H=2+49-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)\)
\(-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+..+\dfrac{50-49}{49.50}\right)\)
\(H-51>-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(H-51>-\left(1-\dfrac{1}{50}\right)\)
\(H>-\dfrac{49}{50}+51>50\)
29 tháng 7 2017
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)( đpcm )