Rút gọn: 1+3+3^2+3^3+.....+3^100
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a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
![](https://rs.olm.vn/images/avt/0.png?1311)
1 + 3 + 32 + 33 + 34 + ........ + 3100
\(3S=3+3^2+3^3+3^4+3^5+.......+3^{101}\)
\(3S-S+\left(3+3^2+3^3+3^4+3^5+.......+3^{101}\right)-\left(1+3+3^2+3^3+3^4+........+3^{100}\right)\)
\(2S=3+3^2+3^3+3^4+3^5+.......+3^{101}-1-3-3^2-3^3-3^4-......-3^{100}\)
\(2S=3^{101}-1\)
\(S=\frac{3^{101}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có:
\(\left(-3\right)^n=3^n\) nếu n chẵn
\(\left(-3\right)^n=-3^n\) nếu n lẻ
B = C - D trong đó
\(C=1+3^2+3^4+...+3^{100}\)
\(D=3+3^3+3^5+...+3^{99}\)
+ \(3C=3+3^3+3^5+...+3^{101}\)
\(2C=3C-C=3^{101}-1\Rightarrow C=\frac{3^{101}-1}{2}\)
+ \(3D=3^2+3^4+3^6+...+3^{100}\)
\(2D=3D-D=3^{100}-3\Rightarrow D=\frac{3^{100}-3}{2}\)
=> \(B=C-D=\frac{\left(3^{101}-1\right)-\left(3^{100}-3\right)}{2}=\frac{\left(3^{101}-3^{100}\right)+2}{2}=\frac{3^{100}\left(3-1\right)+2}{2}=\frac{2\left(3^{100}+1\right)}{2}=3^{100}+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 = \(\frac{3^{101}+1}{4}\)
![](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 +...+3^3 -3^2 +3
=> 4A = 3A + A = 3^101 + 1
A = 3101 + 1
4
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\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\\ \Rightarrow3A=3+3^2+3^3+...+3^{100}+3^{101}\\ \Rightarrow3A-A=3^{101}-1\\ \Rightarrow2A=3^{101}-1\\ \Rightarrow A=\left(3^{101}-1\right).\dfrac{1}{2}\\ \Rightarrow\dfrac{3^{101}}{2}-\dfrac{1}{2}.\)
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
Ta có: \(3A=3+3^2+3^3+...+3^{99}+3^{100}\)
Khi đó: \(3A-A=3+3^2+3^3+...+3^{99}+3^{100}+3^{101}-\left(1+3+3^2+3^3+...+3^{99}+3^{100}\right)\)
\(=3^{101}-1\)
\(\Leftrightarrow2A=3^{101}-1\)
Vậy \(A=\left(3^{101}-1\right):2\)
Ta có: 1+3+32+33+......+3100 (1)
=> gấp tổng lên 3 lần ta có: 3+32+33+34+.....+3101 (2)
Lấy (2)-(1)
TA có: (3+32+33+34+......+3101)-(1+3+32+33+....+3100)
=> 2 lần tổng đó= 3101-1
=> Tổng trên bằng: 3101-1/2