C = 1/3^1+2/3^2+...+100/3^100
Chứng minh C < 3/4
giúp mk nha các bn
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VT
1
31 tháng 1 2017
C = 1/3^1 + 2/3^2 + .3/3^3 + .. + 100/3^100
1/3 . C = 1/3^2 + 2/3^3 + 3/3^4 + .. + 100/3^101
=> C - 1/3 . C = 1/3^1 + (2/3^2 - 1/3^2) + (3/3^3 - 2/3^3) + ... +(100/3^100 - 99/3^100) - 100/3^101
=> 2/3. C = 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 - 100/3^101
xét S= 1/3^1 + 1/3^2 + 1/3^3 + .. + 1/3^100 tương tự
1/3 . S = 1/3^2 + 1/3^3 + 1/3^4 + .. + 1/3^101
=> S - 1/3. S = 1/3^1 - 1/3^101
=> 2/3. S = (1/3 - 1/3^101)
=> S = 3/2. (1/3 - 1/3^101) thay vào C ta có
2/3. C = 3/2. (1/3 - 1/3^101) - 100/3^101
=> C = 9/4. (1/3 - 1/3^101) - 150/3^101
=> C = 3/4 - 9/4*1/3^101 - 150/3^101 < 3/4
TL
1
10 tháng 10 2015
Ta co :
1/2! +2/3! +3/4! +... + 99/100!
= (1/1! -1/2!) + (1/2! - 1/3!) + (1/3! -1/4!) + .... + (1/99! -1/100!)
=1 - 1/100! <1
lik e nhe
LA
12 tháng 3 2019
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
8 tháng 8 2017
\(B=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\)
\(B=\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}\right)+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B=\dfrac{49}{36}+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B=\dfrac{1}{36}+\dfrac{4}{3}+\left(\dfrac{1}{4^2}+...+\dfrac{1}{25^2}\right)\)
\(B>\dfrac{4}{3}\left(1\right)\)
\(\)\(B< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{24.25}\)
\(B< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(B< 2-\dfrac{1}{25}\)
\(B< 2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\dfrac{4}{3}< B< 2\)
\(\rightarrowđpcm\)
26 tháng 7 2017
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
26 tháng 7 2017
1.
B = 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1
3B = 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3
3B + B = ( 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3 ) + ( 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1 )
4B = 3101 + 1
B = \(\frac{3^{101}+1}{4}\)