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\(A=1+2+2^2+...+2^{51}\)
\(2A=2+2^2+2^3+...+2^{52}\)
\(2A-A=\left(2+2^2+2^3+...+2^{52}\right)-\left(1+2+2^2+...+2^{51}\right)\)
\(A=2^{52}-1\)
\(B=5+5^2+5^3+...+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{101}\)
\(5B-B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
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.
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
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) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)
\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)
\(\Rightarrow B+3B=3^{101}-3\)
\(B\left(1+3\right)=3^{101}-3\)
\(4B=3^{101}-3\)
\(B=\frac{3^{101}-3}{4}\)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
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 = \(\frac{2^{103}-2}{3}\)
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 ^98 -2^ 97+....+2 ^2 -2
=>3A=2^ 201 -2
=>A=\(\frac{2^{201}-2}{3}\)
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
=>3B+B=3^ 101 -3^100+3^ 99 -3 ^98+...+3 ^3 -3 ^2+3+3 ^100 -3^ 99+3^ 98 -3^ 97+....+3 ^2 -3+1
=>4B=3 ^101+1
=>B=\(\frac{3^{101}+1}{4}\)
Gọi tổng trên là A , ta có:
A=2+22+23+.....+299+2100
2A=22+23+24+.....+2100+2101
2A-A=(22+23+24+....+2100+2101)-(2+22+23+.....+299+2100)
A=2101-2
Đặt A = 2 + 22 + 23 + .... + 299 + 2100
=> 2A = 22 + 23 + .... + 2100 + 2101
=> 2A - A = 2101 - 2
=>A = 2101 - 2