Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
phần b tương tự phần a nên em làm câu a và c thôi :
a, \(M=1-2+2^2-2^3+...+2^{2012}\)
\(2M=2-2^2+2^3-2^4+...+2^{2013}\)
\(3M=2^{2013}+1\)
\(M=\frac{2^{2013}+1}{3}\)
c, \(E=2^{100}-2^{99}-2^{98}-...-1\)
\(E=2^{100}-\left(2^{99}+2^{98}+...+1\right)\)
đặt \(A=2^{99}+2^{98}+...+1\)
\(2A=2^{100}+2^{98}+...+2\)
\(2A-A=2^{100}-1\) hay \(A=2^{100}-1\)
ta có :
\(E=2^{100}-\left(2^{100}-1\right)\)
\(E=2^{100}-2^{100}+1=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
1.
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 = \(\frac{3^{101}+1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c.
C= ( a+b+c)2+(a+b-c)2- 2(a+b)2
=a2+b2+c2+a2+b2- c2-2a2-2b2
= 2a2+2b2+c2-c2-2a2-2b2
= 0
Vậy C= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
3A = \(3^{101}-3^{100}+3^{99}-3^{98}+...-3^2+3\)
3A + A = \(3^{101}-3^{100}+3^{99}-3^{98}+....-3^2+3+3^{100}-3^{99}+3^{98}-...+3^2-3+1\)
4A = \(3^{101}+1\)
=> A = \(\frac{3^{101}+1}{4}\)
Tock đúng nah
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt 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 + 1
3A +A = 3^101 - 3^100 + 3^99 - 3^98 +.. + 3^3 - 3^2 + 3 + 3^100 - 3^99 + 3^98 -...+3^2 - 3 + 1
4A = 3^101 + 1
A = (3^101 + 1) /4
a) A = 1+3+32+33+...+399+3100
=> 3A=3+32+33+34+...+3100+3101
=> 3A-A=(3+32+33+...+3100+3101)-(1+3+32+33+...+399+3100)
=> 2A=3101-1
=> A = \(\dfrac{3^{101}}{2}\)
b) B=2100-299+298-297+...-23+22-2+1
=> 2B=2101-2100+299-298+...-24+23-22+2
=>2B+B=(2101-2100+299-298+...-24+23-22+2)+(2100-299+298-297+...-23+22-2+1)
=> 3B = 2101+1
=> B = \(\dfrac{2^{101}+1}{3}\)
câu a thì
=> 2A= 3101-1
thì => A=3101-1/2 chứ