3 mũ 1+3mũ2+3mũ3+3mũ4+...3mũ199
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LN
1
NH
1
LP
0
22 tháng 9 2020
Đặt \(D=3-3^2+3^3-3^4+...+3^9-3^{10}+3^{11}\)
=> \(3D=3^2-3^3+3^4-3^5+...+3^{10}-3^{11}+3^{12}\)
Cộng vế 2 BT trên ta được:
\(D+3D=\left(3-3^2+...+3^{11}\right)+\left(3^2-3^3+...+3^{12}\right)\)
\(\Leftrightarrow4D=3^{12}+3\)
\(\Rightarrow D=\frac{3^{12}+3}{4}\)
LT
0
LN
0
NP
4
28 tháng 3 2019
\(Q=1+3+3^2+3^3+3^4+...+3^{11}\)
\(3Q=3+3^2+3^3+3^4+3^5+...+3^{12}\)
\(3Q-Q=\left(3+3^2+3^3+3^4+3^5+...+3^{12}\right)-\left(1+3+3^2+3^3+3^4+...+3^{11}\right)\)
\(2Q=3^{12}-1\)
\(Q=\frac{3^{12}-1}{2}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 10 2018
a/
\(3S=3+3^2+3^3+3^4+...+3^{120}\)
\(2S=3S-S=3^{120}-1\Rightarrow S=\frac{3^{120}-1}{2}\)
b/ \(S=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(S=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(S=13+3^3.13+...+3^{117}.13=13\left(1+3^3+...+3^{117}\right)\) chia hết cho 13
c/
\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(S=40+3^4.40+...+3^{116}.40=40\left(1+3^4+...+3^{116}\right)\) chia hết cho 40
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
Đặt \(A=3^1+3^2+3^3+...+3^{199}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)
Lấy 3A trừ A theo vế ta có :
\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)
\(2A=3^{200}-1\)
\(A=\frac{3^{200}-1}{2}\)
Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)