A=1/5*1*2 + 1/5*1*2*3 + 1/5*1*2*3*4 + .....+ 1/5*1*2*3...*100
Những câu hỏi liên quan
1. (1+1/2).(1+1/2^2).(1+1/2^3)....(1+1/2^100) < 3
2. 1/(5+1)+2/(5^2+1)+4/(5^4+1)+...+ 1024/(5^1024+1) <1/4
3. 3/(1!+2!+3!)+4/(2!+3!+4!)+...+100/(98!+99!+100!) <1/2
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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!
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1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow A< 1\)
Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)
\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)
*Cách khác:
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)
\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Ta thấy:
\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)
\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)
Mà \(\frac{2+1}{2}< 3\)
\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)
\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)
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bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
B = 1 + 5 + 52 + 53 + ....... + 52008 + 52009
S = 1 + 2 + 5 + 14 + ....... + 3n-1 + 1/2 (với n thuộc Z)
A = 1 + 3/2^3 + 4/2^4 + 5/2^5 + ...... + 100/2^100
Q = 1 + 1/2*(1+2) + 1/3*(1+2+3) + 1/4*(1+2+3+4) + ...... + 1/20*(1+2+3+.....+20)
M = -4/1*5 - 4/5*9 - 4/9*13 - ....... - 4/(n+4)*n
Giúp mk với! Mk đang cần gấp lắm !!!!!
\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
Trừ theo vế:
\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
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\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)
\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)
\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)
Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)
\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)
Trừ theo vế:
\(3X-X=3^n-3^0=3^n-1\)
\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
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\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\Rightarrow 2A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
Trừ theo vế:
\(2A-A=1+\frac{3}{2^2}+\frac{4-3}{2^3}+\frac{5-4}{2^4}+\frac{6-5}{2^5}+...+\frac{100-99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Leftrightarrow A=1+\frac{3}{4}-\frac{100}{2^{100}}+(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}})\)
Đặt \(T=(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}})\)
\(\Rightarrow 2T=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\)
Trừ theo vế: \(2T-T=\frac{1}{2^2}-\frac{1}{2^{99}}\)
\(\Leftrightarrow T=\frac{1}{4}-\frac{1}{2^{99}}\)
Do đó: \(A=1+\frac{3}{4}-\frac{100}{2^{100}}+\frac{1}{4}-\frac{1}{2^{99}}=2-\frac{102}{2^{100}}\)
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Tính C=1/2-(1/3+2/3)+(1/4+2/4+3/4)-(1/5+2/5+3/5+4/5)+...+(1/100+2/100+...+99/100)
Tính
A=1/2+1/2^2+1/2^3+...+1/2^100
Tính
B=1/2+1/2^2+1/2^3+1/2^4+...+1/2^99 - 1/2^100
Tính
C=1/2+1/2^3+1/2^5+...+1/2^99
Tính
D=2/3+8/9+26/27+...+3^n-1/3^n.Chứng minh A>n-1/2
Tính: E=4/3+10/9+28/27+...+3^39+1/3^92.Chứng minh B<100
Tính
F=5/4+5/4^2+5/4^3+...+5/4^99.Chứng minh C<5/3
Tính
G=3/1^2*2^2+5/2^2*3^2+7/3^2*4^2+...+19/9^2*10^2.Chứng Minh D<1
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
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Chứng minh rằng:a. frac{1}{3^2}+frac{2}{3^3}+frac{3}{3^4}+frac{4}{3^5}+...+frac{99}{3^{100}}+frac{100}{3^{101}} frac{1}{4}b.frac{1}{2}-frac{1}{4}+frac{1}{8}-frac{1}{16}+frac{1}{32}-frac{1}{64} frac{1}{3}c.frac{1}{3}-frac{2}{3^2}+frac{3}{3^3}-frac{4}{3^4}+...+frac{99}{3^{99}}-frac{100}{3^{100}} frac{1}{16}d. frac{1}{5^2}-frac{2}{5^3}+frac{3}{5^4}-frac{4}{5^5}+...+frac{99}{5^{100}}-frac{100}{5^{101}} frac{1}{36}
Đọc tiếp
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Tính nhanh:
A = 1/3 - 3/4 - ( - 3/5 ) + 1/72 - 2/9 - 1/36 + 1/15
B = 1/ 5 - 3/7 + 5/9 - 2/11 + 7/13 - 9/16 - 7/13 + 2/11 - 5/9 + 3/7 - 1/5
C = 1/100 - 1/100 . 99 - 1/99 . 98 - 1/98 . 97 - ... - 1/3 . 2 - 1/ 2 . 1
C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
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C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
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\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)
\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(=0+0+0+0-\frac{1}{16}\)
\(=\frac{-1}{16}\)
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a) Rút gọn biểu thức sau:
A=2*2^2+3*2^3+4*2^4+5*2^5+...+100*2^100
b) Cho B=1/2-1/3+1/4-1/5+...+1/98-1/99
CMR: 0,2< B < 0,4
cho M = 1/3 + 1/5 + 1/9 + 1/17 + 1/33 + 1/65 va N = 1/5^2 - 2/5^3 + 3/5^4 - 4/5^5 + . . . + 99/5^100 - 100/5^101
Chứng minh rằng
D= 1/3-2/32+3/33-4/34+..........+99/399-100/3100<3/16
E=1/52-2/53+3/54-4/55+.......+99/5100-100/5101<1/36
F=1/22+1/32+1/42+.......+1/502<1