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A=(2^1+2^2)+(2^3+2^4)+.....+(2^99+2^100)
A=(2+2^2)+2^2(2+2^2)+.....+2^98(2+2^2)
A=6+2^2.6+....+2^98.6
A=6+2^2.6+......+2^98.3.2
Vậy A chia hêt cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
A = 2 + 22 + 23 + 24 + .... +299 + 2100
2A = 22 + 23 + 24 + 25 + ... + 2100 + 2101
2A - A = A = 2101 - 2
vậy A = 2101 - 2
b)
B = 1 + 2 + 22 + 23 + ... + 22017
2B = 2 + 22 + 23 + 24 + ... + 22018
2B - B = B = 22018 - 1
Vậy B = 22018 - 1
c)
C = 2 + 23 + 25 + ... + 22017
4C = 23 + 25 + 27 + ... + 22019
4C - C = 3C = 22019 - 2
C = \(\frac{2^{2019}-2}{3}\)
d)
D = 2 + 24 + 27 + ... + 22017
8D = 24 + 27 + 210 + ... + 22020
8D - D = 7D = 22020 - 2
D = \(\frac{2^{2020}-2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trước tiên ta tính
B = 1*4 + 4*7 + 7*10 + 10*13 + ... + 100*103
9B = 1*4*(7+2) + 4*7*(10-1) + 7*10*(13-4) + 10*13*(16-7) +...+100*103*(106-97)
9B = 1*2*4 + 1*4*7 - 1*4*7 + 4*7*10 - 4*7*10 + 7*10*13 -7*10*13+10*13*16 + ...+ 100*103*106 - 97*100*103
9B = 1*2*4 + 100*103*106 => B = 121312
Sau đó ta tính C
C = 1 + 4 + 7 +10 + ... + 100 = \(\frac{1}{2}\left(100+1\right)\left(\frac{\left(100-1\right)}{3}+1\right)=\frac{1}{6}101\cdot102\)=1717
Và: B - 3C =
B = 1*4 + 4*7 + 7*10 + 10*13 + ... + 100*103
-3C= -1*3 - 4*3 - 7*3 - 10*3 - ... - 100*3
B-3C = 12 + 42 + 72 + 102 + ...+ 1002 =A = 121312 - 3*1717 = 116161.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2+2^2+2^3+2^4+.....+2^{98}+2^{99}\)
\(\Rightarrow2A=2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\right)-\left(2+2^2+2^3+2^4+.....+2^{98}+2^{99}\right)\)
\(\Rightarrow A=2^{100}-2\)
b) \(B=2+2^4+2^7+......+2^{97}+2^{100}\)
\(\Rightarrow2^3B=2^4+2^7+......+2^{100}+2^{103}\)
\(\Rightarrow8.B-B=\left(2^4+2^7+......+2^{100}+2^{103}\right)-\left(2+2^4+2^7+......+2^{97}+2^{100}\right)\)
\(\Rightarrow7B=2^{103}-2\)
\(\Rightarrow B=\dfrac{2^{103}-2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2+22+23+....+210
=> 2A=2(1+2+22+23+....+210)
=> 2A=2+22+23+24+....+211
=> 2A-A=211-1
*) B=1+3+32+33+....+3100
=> 3B=3+32+33+34+....+3101
=> 2B=3101-1
=> B=\(\frac{3^{101}-1}{2}\)
2A= 2 + 22 + 24 + 25 + 26 + 27 + 28 +29 + 210 + 211
Lay 2A- A = (2 + 22 + 24 + 25 + 26 + 27 + 28 +29 + 210 + 211) - ( 1 + 2 + 22 + 24 + 25 + 26 + 27 + 28 +29 + 210)
A = 211-1
3B = 3 + 32 + 33 + 34 + 35 + ... + 3101
Lay 3B - B = (3 + 32 + 33 + 34 + 35 + ... + 3101) - ( 1 + 3 + 32 + 33 + 34 + ... + 3100)
2B = 3101-1
B = (3101-1)/ 2
đề bài