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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
a) A = 20 + 21 + 22 + .... + 22010
2A = 2(20 + 21 + 22 + .... + 22010)
2A = 21 + 22 + 23 + .... + 22011
A = (21 + 22 + 23 + .... + 22011) - (20 + 21 + 22 + .... + 22010)
A = 22011 - 20
A = 22011 - 1
b) B = 1 + 3 + 32 + .... + 3100
3B = 3(1 + 3 + 32 + .... + 3100)
3B = 3 + 32 + 33 + .... + 3101
2B = (3 + 32 + 33 + .... + 3101) - (1 + 3 + 32 + .... + 3100)
2B = 3101 - 1
B = (3101 - 1) : 2
c) C = 4 + 42 + 43 + .... + 4n
4C = 4(4 + 42 + 43 + .... + 4n)
4C = 42 + 43 + 44 .... + 4n + 1
3C = (42 + 43 + 44 .... + 4n + 1) - (4 + 42 + 43 + .... + 4n)
3C = 4n + 1 - 4
C = (4n + 1 - 4) : 3
d) D = 1 + 5 + 52 + .... + 52000
5D = 5(1 + 5 + 52 + .... + 52000)
5D = 5 + 52 + 53 + .... + 52001
4D = (5 + 52 + 53 + .... + 52001) - (1 + 5 + 52 + .... + 52000)
4D = 52001 - 1
4D = (52001 - 1) : 4
Bài 1
a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)
\(=-\dfrac{24}{33}\)
a) >
b) >
c) <
d) =