Cho M= 1/1+2+3 + 1/1+2+3+4 + ... +1/1+2+3+...+59
CMR: M<2/3
Những câu hỏi liên quan
Cho 1/M=1/(1+2+3) + 1/(1+2+3+4) +.....+ 1/(1+2+3+4+...+59)
Chứng minh rằng M>2/3
1) Cho
1/M=1/1+2+3+1/1+2+3+4+..+1/1+2+3+..+59
CMR M>2/3
phải là M<2/3 mới giải đc
\(M=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+59}\)
\(=\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(59+1\right).59:2}\)
\(=\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1770}\)
\(=\frac{2}{12}+\frac{2}{20}+...+\frac{2}{3540}\)
\(=2\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{595.60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)
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Cho M=(1/1+2+3)+(1/1+2+3+4)+...+(1/1+2+3+...+59) . Chứng minh M<2/3
Bài 1:
a) C/m: A=2^1+2^2+2^3+2^4+....+2^2010 chia het cho 3 và 7
b) C/m: B=3^1+3^2+3^3+3^4+....+3^2010 chia het cho 4 va 13
c) C/m: C= 5^1+5^2+5^3+5^4+....+5^2010 chia het cho 6 va 31
d) C/m: D=7^1+7^2+7^3+7^4+....+7^2010 chia het cho 8 va 57
1/M=1/1+2+3+1/1+2+3+4+1/1+2+3+4+5+...+1/1+2+3+4+..+59
cmr M>2/3
1. Cho Ndfrac{1}{31}+dfrac{1}{32}+...+dfrac{1}{60}CMR dfrac{3}{5} N dfrac{4}{5}2. Cho Mdfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+...+dfrac{29}{3^{29}}-dfrac{30}{3^{30}}CMR M dfrac{3}{16}3. Cho Qdfrac{2}{3}+dfrac{8}{9}+dfrac{26}{27}+...+dfrac{3^{2021}-1}{3^{2021}}CMR Qdfrac{4041}{2}
Đọc tiếp
1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
cho M=1/(1*2*3)+1/(2*3*4)+...+1/[n(n+1)(n+2)]
va N=[n(n+3)/[4(n+1)(n+2)]
tinh M-N
cho biểu thức M=2^3-1/2^3+1x3^3-1/3^3+1x4^3-1/4^3+1x...x100^3-1/100^3+1.Chứng minh rằng M>2/3
1) Cho
1/M=1/1+2+3+1/1+2+3+4+..+1/1+2+3+..+59
CMR M>2/3
2)Tính:
P=(1-1/111)(1-2/111)...(1-2009/111)
P=\(\left(1-\dfrac{1}{111}\right)\left(1-\dfrac{2}{111}\right)\times...\times\left(1-\dfrac{111}{111}\right)\times...\times\left(1-\dfrac{2009}{111}\right)\)
P=\(\left(1-\dfrac{1}{111}\right)\left(1-\dfrac{2}{111}\right)\times...\times0\times...\times\left(1-\dfrac{2009}{111}\right)\)
P=0
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