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1/2^2+1/3^2+1/4^2+...+1/100^2
+) 1/2^2=1/2.2< 1/1.2
+) 1/3^2 = 1/3.3 < 1/2.3
+) 1/4^2 =1/4.4 < 1/3.4
+) ...
+) 1/100^2 = 1/100.100 < 1/99.100
=> 1/2^2+1/3^2+1/4^2+...+1/100^2 < 1/1.2+ 1/2.3+1/3.4+..+1/99.100 = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 = 1-1/100 < 1
=> 1/2^2+1/3^2+1/4^2+...+1/100^2 < 1
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a, \(2+2^2+.....+2^{49}+2^{50}=2^{1+2+..+50}=2^{\frac{\left(50+1\right)\left[\left(50-1\right):1+1\right]}{2}}=1275\)
b, tương tự
Ta có A<1/12+1/1.2+1/2.3+1/3.4+...+1/49.50
A<1+1-1/2+1/2-1/3+1/3-1/4+1/49-1/50
A<2-1/50<2(đpcm)
Chuyển đổi \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....=\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-.......+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+.......+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)
\(A=\frac{1}{1}-0+0+0+0+.......+0+0-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Có \(\frac{49}{50}<2\) nên \(A<2\)
GIẢI
A=1/12 +1/22 + 1/32 + ....+ 1/502
A<1+1/1.2+1/2.3+....+1/49.50
A<1+1-1/2+1/2-1/3+...+1/49-1/50
A<2-1/50<2
vậy A<2
\(A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49x50}=1-\frac{1}{50}\)
\(=>A<1-\frac{1}{50}<2\)
\(=>A<2\)
Ta có:
\(\frac{1}{1^2}=1;\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{50^2}<\frac{1}{49.50}\)
=>A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<1+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}<2\)
=> A < 2
k nha
1/22 < 1/1.2
1/32<1/2.3
......
1/502<1/49.50
\(\Rightarrow\)1/22+1/32+.....+1/502<1/1.2+1/2.3+.........+1/49.50
\(\Rightarrow\)1/22+1/32+.....+1/502<1/1-1/2+1/2-1/3+......+1/49-1/50
\(\Rightarrow\)1/22+1/32+.....+1/502<1/1-1/50
\(\Rightarrow\)1/22+1/32+.....+1/502<49/50
\(\Rightarrow\)1/22+1/32+.....+1/502 +1<49/50 +1
\(\Rightarrow\)A<\(1\frac{49}{50}\)
Vì \(1\frac{49}{50}<2\)
\(\Rightarrow\)A<2