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\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)
Sửa đề: \(S=2^{100}-2^{99}+2^{98}-...+2^2-2\)
=>\(2\cdot S=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
=>\(2S+S=2^{100}-2^{99}+2^{98}-...+2^2-2+2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
=>\(3S=2^{101}-2\)
=>\(S=\dfrac{2^{101}-2}{3}\)
a: \(A=3^{100}-3^{99}+3^{98}-...+3^2-3\)
=>\(3A=3^{101}-3^{100}+3^{99}-...+3^3-3^2\)
=>\(4A=3^{101}-3\)
=>\(A=\dfrac{3^{101}-3}{4}\)
b: \(B=\left(-2\right)^0+\left(-2\right)^1+...+\left(-2\right)^{2024}\)
=>\(B\cdot\left(-2\right)=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}\)
=>\(-2B-B=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}-\left(-2\right)^0-\left(-2\right)^1-...-\left(-2\right)^{2024}\)
=>\(-3B=-2^{2025}-1\)
=>\(B=\dfrac{2^{2025}+1}{3}\)
c: \(C=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2023}\)
=>\(\left(-\dfrac{1}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2024}\)
=>\(\left(-\dfrac{6}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^{2024}-\left(-\dfrac{1}{5}\right)^0\)
=>\(C\cdot\dfrac{-6}{5}=\dfrac{1}{5^{2024}}-1=\dfrac{1-5^{2024}}{5^{2024}}\)
=>\(C\cdot\dfrac{6}{5}=\dfrac{5^{2024}-1}{5^{2024}}\)
=>\(C=\dfrac{5^{2024}-1}{5^{2024}}:\dfrac{6}{5}=\dfrac{5^{2024}-1}{6\cdot5^{2023}}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
297 . 299
= 297 . ( 298 + 1 )
= 297 . 298 + 297
2982 = 298 . 298
= ( 297 + 1 ) . 298
= 297 . 298 + 298
Mà 297 . 298 + 297 < 297 . 298 + 298 nên 297 . 299 < 2982 ( đpcm )
Đặt :
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)
Vậy..
\(A=1-2+3-4+5-6+7-8+...+99-100\)
\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(A=\left(-1\right).50\)
\(A=-50\)
\(B=1+3-5-7+9+11-...-397-399\)
\(B=1-2+2-2+2-...+2-2-399\)
\(B=1-399\)
\(B=-398\)
\(C=1-2-3+4+5-6-7+...+97-98-99+100\)
\(C=-1+1-1+1-...-1+1\)
\(C=0\)
\(D=2^{2024}-2^{2023}-...-1\)
\(D=2^{2024}-\left(2^0+2^1+2^2+...2^{2023}\right)\)
\(D=2^{2024}-\left(\dfrac{2^{2024}-1}{2-1}\right)\)
\(D=2^{2024}-\left(2^{2024}-1\right)\)
\(D=2^{2024}-2^{2024}+1\)
\(D=1\)
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +...+ 99 - 100
A = (1 - 2) + ( 3 - 4) + ( 5- 6) +....+(99 - 100)
Xét dãy số 1; 3; 5;...;99
Dãy số trên là dãy số cách đều có khoảng cách là: 3 - 1 = 2
Dãy số trên có số số hạng là: (99 - 1) : 2 + 1 = 50 (số)
Vậy tổng A có 50 nhóm, mỗi nhóm có giá trị là: 1- 2 = -1
A = - 1\(\times\)50 = -50
b,
B = 1 + 3 - 5 - 7 + 9 + 11-...- 397 - 399
B = ( 1 + 3 - 5 - 7) + ( 9 + 11 - 13 - 15) + ...+( 393 + 395 - 397 - 399)
B = -8 + (-8) +...+ (-8)
Xét dãy số 1; 9; ...;393
Dãy số trên là dãy số cách đều có khoảng cách là: 9-1 = 8
Dãy số trên có số số hạng là: ( 393 - 1): 8 + 1 = 50 (số hạng)
Tổng B có 50 nhóm mỗi nhóm có giá trị là -8
B = -8 \(\times\) 50 = - 400
c,
C = 1 - 2 - 3 + 4 + 5 - 6 +...+ 97 - 98 - 99 +100
C = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7+ 8) +...+ ( 97 - 98 - 99 + 100)
C = 0 + 0 + 0 +...+0
C = 0
d, D = 22024 - 22023- ... +2 - 1
2D = 22005- 22004 + 22003+...- 2
2D + D = 22005 - 1
3D = 22005 - 1
D = (22005 - 1): 3
SSH: (399-1):2+1= 200
Neu chia moi nhom 4 so thi so cap so la:
200:4 = 50
Ta co:
B=1+3-5-7+9+11-13-15+...+393+395-397-399
B= (1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)
B= -8 + -8 +...+ -8
B= -8 . 50
B= -400
SSH: (399-1):2+1= 200
Neu chia moi nhom 4 so thi so cap so la:
200:4 = 50
Ta co:
A=1+3-5-7+9+11-13-15+...+393+395-397-399
A= (1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)
A= -8 + -8 +...+ -8
A= -8 . 50
A= -400
a) \(A=1+2+2^2+2^3+...+2^{99}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{100}-1-2-2^2-...-2^{99}=2^{100}-1\)
b) \(A=1+2+2^2+...+2^{99}=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)
\(=15+2^4.15+...+2^{96}.15=15\left(1+2^4+...+2^{96}\right)\)
\(=3.5\left(1+2^4+...2^{96}\right)\) chia hết cho 3 và 5
c) \(A=1+2+2^2+...+2^{99}\)
\(=1+2\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)
\(=1+2.7+...+2^{97}.7=1+7\left(2+...+2^{97}\right)\) chia 7 dư 1
=> A không chia hết cho 7
A=2100-299+298-297+.....+22-2
=>2A=2101-2100+299-298+.....+23-22
=>2A+A=2101-2100+299-298+.....+23-22+2100-299+298-297+....+22-2
=>3A=2201-2
=>A=\(\frac{2^{201}-2}{3}\)
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=\(\frac{3^{101}+1}{4}\)
Câu a : Đặt 2A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2
=> 2A - A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - ( 2^100 - 2^99 + 2^98 - 2^97 +...+ 2^2 - 2)
=> A = 2^101 - 2^100 + 2^99 - 2^98 +...+ 2^3 - 2^2 - 2^100 + 2^99 - 2^98 + 2^97 -...- 2^2 + 2
=> A= = 2^101 -2(2^100 + 2^98 + 2^96 +...+ 2^2) + 2(2^99 + 2^97 + 2^95 +...+ 2^3) +2
Câu b : Làm tương tự như trên
BẤM ĐÚNG CHO MÌNH NHA