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Câu 1:
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)
\(A=\frac{1}{1\times1}+\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+.....+\frac{1}{50\times50}\)
\(A< \frac{1}{1\times1}+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....+\frac{1}{49\times50}\)
\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(A< 2-\frac{1}{50}< 2\)
Câu 2:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+.....+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+.....+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+.....+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+.....+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{3069}{512}\)
Câu 3:
\(\frac{1}{2\times3}=\frac{1}{6}\)
\(\frac{1}{2}-\frac{1}{3}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\)
\(\Rightarrow\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
Câu 4:
\(M=\frac{9}{40}-\frac{11}{60}+\frac{13}{84}-\frac{15}{112}\)
\(M=\left(\frac{9}{40}-\frac{11}{60}\right)+\left(\frac{13}{84}-\frac{15}{112}\right)\)
\(M=\left(\frac{27}{120}-\frac{22}{120}\right)+\left(\frac{52}{336}-\frac{45}{336}\right)\)
\(M=\frac{1}{24}+\frac{1}{48}\)
\(M=\frac{2+1}{48}\)
\(M=\frac{3}{48}\)
\(M=\frac{1}{16}\)
Chúc bạn học tốt
câu 2:
s= 3+3/2+3/3^2+.....+3/2^9
=> 2s=6+3+3/2+...+3/2^8
=> 2s-s =( 6+3+3/2 + ....+3/2^8)- ( 3+3/2 +3/2^2+...+3/2^9)
=> s=6-3/2^9=3069/512
Bài 1:
\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1+1-\dfrac{1}{50}\)
\(=2-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
Vậy...
A=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)
A=1+\(\frac{1}{2^2}\)\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)
A<1+\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{49\cdot50}\)
A<1+1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A<2-\(\frac{1}{50}\)<2
=>A<1(câu 1)
A= \(\dfrac{1}{1^2}\)