DT
1
K
Khách
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ND
1
LB
14 tháng 5 2018
DẶT A= BIỂU THỨC TRÊN
A=2+1+1+..+1-(1/4+1/9+...+1/2500)
ĐẶT S=1/4+1/9+...+1/2500
S=1/2^2+1/3^2+...+1/50^2
SÓ SỐ HẠNG CỦA S:
(50-2)/1+1=49
SUY RA
1+1+...+1=49
SUY RA A=2+49-S
A=51-S
TAO CÓ :
S<1/1.2+1/2.3+...+1/49.100
S<1-1/2+1/2-1/3+...+1/49-1/50
S<1-1/50
S<49/50
SUY RA A>51-49/50
SUY RA A>50
NT
0
NM
1
XO
27 tháng 8 2020
\(C=1+1+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
51 số hạng 49 số hạng
= \(51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)\)
\(>51-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)=51-\left(\frac{1}{2}-\frac{1}{51}\right)=51-\frac{1}{2}+\frac{1}{51}\)
\(=50,5+\frac{1}{51}>50\left(đpcm\right)\)
Vậy C > 50
KT
1
LB
4 tháng 5 2018
\(H=2+\dfrac{4-1}{4}+\dfrac{9-1}{9}+\dfrac{16-1}{16}+..+\dfrac{2500-1}{2500}\)\(H=2+49-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)\)
\(-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+..+\dfrac{50-49}{49.50}\right)\)
\(H-51>-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(H-51>-\left(1-\dfrac{1}{50}\right)\)
\(H>-\dfrac{49}{50}+51>50\)
T
0
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2021
\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)
Từ đó ta có:
\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{50^2-1}{50^2}>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+...+1-\dfrac{1}{49.50}\)
\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)
16 tháng 5 2021
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)
Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)
\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)
\(A=2+\frac{3}{4}+\frac{8}{9}+......+\frac{2499}{2500}\)
\(A=2+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+.....+\left(1-\frac{1}{2500}\right)\)
\(A=2+1-\frac{1}{4}+1-\frac{1}{9}+.........+1-\frac{1}{2500}\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+....+\frac{1}{2500}\right)\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{50^2}\right)\)
Vì mỗi số 1 đều đi với 1 phân số nên có số số 1 là: (50-1)/1+1=50(số)
\(A=52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{50^2}\right)\)
\(\frac{1}{2^2}<\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}\)
.........
\(\frac{1}{50^2}<\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{49}{50}\)
\(\Rightarrow52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)>52-\frac{49}{50}\)
\(\Rightarrow A>51\frac{1}{50}\)
Vì\(51\frac{1}{50}>50\Rightarrow A>50\)