Tính.
a)A=1/1.3.5+1/3.5.7+1/5.7.9+...+1/97.99.101
b)B=2^2019-2^2018-2^2017-...-2+1
c)C=1/1.2+1/3.4+...+1/49.50-1/50-1/49-...-1_2
d)D=1/2!+5/3!+11/4!+..+99.100-1/100!
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Những câu hỏi liên quan
3 tháng 8 2018
A = 1 + 2 + 3 + ... + 2018 (có 2018 số )
= (2018 + 1) . 2018 : 2 = 2037171
B = 1 + 3 + 5 + ... + 2017(có 1009 số )
= (2017 + 1) . 1009 : 2 = 1018081
C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )
= (2018 + 2) x 1009 : 2 = 1019090
D = 72 . 153 + 27.153 + 153
= (72 + 27 + 1) . 153
= 100 . 153 = 15300
13 tháng 10 2023
a) 9 + 99 + 999 + ... + 999999
= (10 - 1) + (100 - 1) + (1000 - 1) + ... + (1000000 - 1)
= (101 + 102 + 103 + ... + 106) - (1.6)
= 1111110 - 6 = 1111104
b) 1 + 11 + 111 + ... + 1111111
= 1 + (101 + 1) + (102 + 101 + 1) + ... + (106 + 105 + 104 + 103 + 102 + 101 + 1)
= 101 . 6 + 102 . 5 + 103 . 4 + ... + 106. 1) + (1 + 1.6)
= 60 + 500 + 4000 + ... + 1000000 + 7
= 1234560 + 7 = 1234567
c) C = 1.2 + 2.3 + 3.4 + 4.5 + ... + 98.99
3C = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 98.99.3
3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97)
3C = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 +...+ 98.99.100 - 98.99.97
3C = 98.99.100
C = \(\dfrac{98.99.100}{3}\) = 323400
d) D = 1.3.5 + 3.5.7 + 5.7.9 + ... + 95.97.99
8D = 1.3.5.8 + 3.5.7.8 + 5.7.9.8 + ... + 95.97.99.8
8D = 1.3.5.(7 + 1) + 3.5.7.(9 - 1) + 5.7.9.(11 - 3) + ... + 95.97.99.(101 - 93)
8D = 1.3.5.7 + 1.3.5.1 + 3.5.7.9 - 3.5.7.1 + 5.7.9.11 - 5.7.9.3 + ... + 95.97.99.101 - 95.97.99.93
8D = 1.3.5.1 + 95.97.99.101
D = \(\dfrac{1.3.5.1+95.97.99.101}{8}=15517600\)
31 tháng 7 2018
a)1.22 + 2.32 + 3.42 + ... + 99.1002
= 1.2(3 - 1) + 2.3(4 - 1) + 3.4(5 - 1) + ... + 99.100(101 - 1)
= 1.2.3 - 1.2 + 2.3.4 - 2.3 + 3.4.5 - 3.4 + ... + 99.100.101 - 99.100
= (1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101) - (1.2 + 2.3 + 3.4 + ... + 99.100)
26 tháng 7 2017
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
26 tháng 7 2017
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
16 tháng 8 2016
A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
A = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
A = B - 2C ( ĐPCM )
Vậy A = B - 2C
23 tháng 8 2019
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
23 tháng 8 2019
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
I
6 tháng 6 2021
1
Ta có :A=1/1.2+1/3.4+...+1/99.100=1/2+1/12+...+1/9900
7/12=1/2+1/12
Vì 1/2+1/12<1/2+1/12+...+1/9900
Nên: 7/12<A (1)
Lại có:A=1/1.2+1/3.4+...+1/99.100
=1-1/2+1/3-1/4+...+1/99-1/100
=(1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
5/6=1-1/2+1/3
vì: 1-1/2+1/3 < (1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
nên 5/6 < A (2)
Từ (1) và (2) suy ra 7/12<A<5/6