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Ta có : a) A= 1+ 5+ 52+ 53+........+ 51998
=> 5A = 5+ 52+ 53+........+ 51999
=> 5A - A = 51999 - 1
=> 4A = 51999 - 1
\(\Rightarrow A=\frac{5^{1999}-1}{4}\)
b) Ta có : b) B= 1+ 4+ 42 + ...... + 4n
=> 4B = 4 + 42 + 43 + ...... + 4n + 1
=> 4B - B = 4n + 1 - 1
=> 3B = 4n + 1 - 1
=> \(B=\frac{4^{n+1}-1}{3}\)
a) S1 = 2.4 + 4.6 + 6.8 + ...+ 100.102
6.S1 = 2.4.6 + 4.6.(8 - 2) + 6.8.(10 - 4) + ...+ 100.102.(104 - 98)
6.S1 = 2.4.6 + 4.6.8 - 2.4.6 + 6.8.10 - 4.6.8 + ....+ 100.102.104 - 98.100.102
6.S1 = (2.4.6 + 4.6.8 + 6.8.10 + ...+ 100.102.104) - (2.4.6 + 4.6.8 + ...+ 98.100.102)
6.S1 = 100.102.104 => S1 = 100.102.104 : 6 = ...
b) S2 = (1 - 2)(1+ 2) + (3 - 4).(3 + 4) + ...+ (55 - 56).(55 + 56) + 572
= (-1).(1 + 2) + (-1).(3 + 4) + ...+ (-1).(55 + 56) + 572 = (-1).(1 + 2+ 3 + 4+...+ 55 + 56) + 572 = -(1+ 56).56 : 2 + 572 = ...
c) S3 = 1.2.( 3 - 1) + 2.3.(4 - 1) + 3.4.(5 - 1) + ....+ 20.21.(22 - 1)
= (1.2.3 + 2.3.4 + 3.4.5 + ...+ 20.21.22) - (1.2 + 2.3 + ...+ 20.21)
Tính A = 1.2.3 + 2.3.4 + 3.4.5 + ...+ 20.21.22
4.A = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5(6 - 2) + ...+ 20.21.22.(23 - 19)
4.A = (1.2.3.4 + 2.3.4.5 + ...+ 20.21.22.23) - (1.2.3.4 + 2.3.4.5 + ....+ 19.20.21.22)
4.A = 20.21.22.23 => A =
Tính B = 1.2 + 2.3 + ...+ 20.21
3.A = 1.2.3 + 2.3.(4 - 1) + ...+ 20.21.(22 - 19) = (1.2.3 + 2.3.4 + ...+ 20.21.22) - (1.2.3+ ...+ 19.20.21) = 20.21.22 => B = ...
d) S4 = 1 + 8 + 27 + 64 + 125 = ....
a)A=1+2+22+...+2100
=>2A=2+22+23+...2101
=>2A-A=(2+22+23+...+2101)-(1+2+22+...+2100)
=>A=2101-1
b)B=3+32+33+...+3100
=>3B=32+33+...+3101
=>3B-B=(32+33+...+3101)-(3+32+...3100)
=>2B-B=3101-3
=>B=(3101-3):2
c)C=1+2+4+8+16+...+8192
=>C=1+2+22+23+...213
=>2C=2+22+23+...+214
=>2C-C=(2+22+...+214)-(2+22+...+213)
=>C=214-2
d)D=4+42+43+...+4n
=>4D=42+43+...+4n+1
=>4D-D=(42+43+...+4n+1)-(4+42+...+4n)
=>3D=4n+1-4
=>D=(4n+1-4):3
13 + 23 + 33 + 43
= 1 + 8 + 27 + 64
= 100 = 102
13 + 23 + 33 + 43 + 53
= 1 + 8 + 27 + 64 + 125
= 225 = 152
Chúc các bạn học tốt
a. 10^2
b. 15^2
tq: 13 + 23 + ... + n3 = (1+...+n)2.
Chúc bạn học tốt.
\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
2; 3 tuong tu
1) A = 1 + 2 + 22 + 23 + .... + 22018
2A = 2 + 22 + 23 + 24 + ..... + 22019
2A - A = ( 2 + 22 + 23 + 24 + ..... + 22019 ) - ( 1 + 2 + 22 + 23 + .... + 22018 )
Vậy A = 22019 - 1
2) B = 1 + 3 + 32 + 33 + ..... + 32018
3A = 3 + 32 + 33 + ...... + 32019
3A - A = ( 3 + 32 + 33 + ...... + 32019 ) - ( 1 + 3 + 32 + 33 + ..... + 32018 )
2A = 32019 - 1
Vậy A = ( 32019 - 1 ) : 2
3) C = 1 + 4 + 42 + 43 + ...... + 42018
4A = 4 + 42 + 43 + ...... + 42019
4A - A = ( 4 + 42 + 43 + ...... + 42019 ) - ( 1 + 4 + 42 + 43 + ...... + 42018 )
3A = 42019 - 1
Vậy A = ( 42019 - 1 ) : 3
a) 23= 8
, 24=16
, 25=32
, 26 =64,
27 =128
28 =256
29 =512
210 =1024
b) 32 =9
33 =27
34 =81
35 =243
c) 42 =16
43 =64
44 =256
d) 52 =25
53 =125
54 =625
e) 62 = 36
63 =216
64 =1296
hok tốt
a, Đặt \(A=\dfrac{3}{1.6}+\dfrac{3}{6.11}+...+\dfrac{3}{496.501}\)
\(5A=\dfrac{3.5}{1.6}+\dfrac{3.5}{6.11}+...+\dfrac{3.5}{496.501}\)
\(5A=3\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{496.501}\right)\)
\(5A=3\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(5A=3\left(1-\dfrac{1}{501}\right)\)
\(5A=3\cdot\dfrac{500}{501}\)
\(A=\dfrac{1500}{501}:5\)
\(A=\dfrac{100}{167}\)
b, Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2018}}\)
\(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\)
\(2B-B=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2018}}\right)\)
\(B=\dfrac{1}{2^2}-\dfrac{1}{2^{2018}}\)