rút gọn c=3^100-3^99-3^98-3^97+.+3^2-3+1
Rút gọn biểu thức
b) B=2^100-2^99+2^98-2^97+...+2^2-2
c) C=3^100-3^99+3^98-3^97+...+3^2-3+1
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
Bài 1 :Rút gọn A=2^100-2^99+2^97+...+2^2 -2
B=3^100-3^99+3^98-3^97+...+3^@ +1
bài 2:
Cho C =1/3+1/3^2+1/3^3+...=1/3^99
CMR C<1/2
C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
Rút gọn B=2^100-2^99+2^98-2^97+...+2^2-2 ; C= 1+(-3)+(-3)^2+...+(-3)^100
Rút gọn A= 3^100 - 3^99 + 3^98 - 3^97 + ... + 3^2 - 3 + 1
Xin lỗi, nhìn nhầm:
A = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1
3A = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3
=> 4A = 3A + A = 3^101 + 1
A = \(\frac{3^{101}+1}{4}\)
B = 3^100 - 3^99 + 3^98 - 3^97 +...........+ 3^2 - 3 + 1
3B = 3^101 - 3^100 + 3^99 - 3^98 +...+3^3 -3^2 +3
Cộng vế với vế triệt tiêu, ta có :
4B = 3^101 + 1
B = \(\frac{3^{101}+1}{4}\)
1.Rút gọn:
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow A+2A=2^{101}-2\)
\(A\left(1+2\right)=2^{101}-2\)
\(A.3=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)
\(\Rightarrow B+3B=3^{101}-3\)
\(B\left(1+3\right)=3^{101}-3\)
\(4B=3^{101}-3\)
\(B=\frac{3^{101}-3}{4}\)
Rút gọn :
\(A=2^{100}-2^{99}+2^{98}-2^{97}.............+2^2-2\)
\(B=3^{100}-3^{99}+3^{98}-3^{97}.........+3^2-3+1\)
a, \(A=...\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\frac{2^{101}-2}{3}\)
b, tương tự a \(B=\frac{3^{101}+1}{4}\)
Rút gọn:
a/ A=2^100-2^99+2^98-2^97+............+2^2-2
b/ B=3^100-3^99+3^98-3^97+..............+3^2-3+1
Ai nhanh nhất là đúng nhất mk tick cho
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
Rút gọn:
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
A = 2100 - 299 + 298 - 297 + ... + 22 - 2
= ( 2100 + 298 + ... + 22 ) - ( 299 + 297 + ... + 2 )
= ( 2100 + 298 + ... + 22 ) - 2( 299 + 297 + ... + 2 ) + ( 299 + 297 + ... + 2 )
= 299 + 297 + ... + 2
=> 4A = 2103 + 299 + ... + 23
=> 3A = 2103 - 2
=> A = \(\frac{2^{103}-2}{3}\)
Rút gọn (1/99+2/98+3/97+...+99/1):(1/2+1/3+1/4+...+1/100)
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^