3mũ1-3mũ2+3mũ3-3mũ4+ . . . +3mũ9-3mũ10+3mũ11
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\(Q=1+3+3^2+3^3+3^4+...+3^{11}\)
\(3Q=3+3^2+3^3+3^4+3^5+...+3^{12}\)
\(3Q-Q=\left(3+3^2+3^3+3^4+3^5+...+3^{12}\right)-\left(1+3+3^2+3^3+3^4+...+3^{11}\right)\)
\(2Q=3^{12}-1\)
\(Q=\frac{3^{12}-1}{2}\)
b, = \(\frac{3^9\cdot3\cdot11+15\cdot3\cdot3^8}{16\cdot3^9}\)
= \(\frac{3^9\cdot33+15\cdot3^9}{16\cdot3^9}\)
= \(\frac{3^9\cdot\left(33+15\right)}{3^9\cdot16}\)
=\(\frac{48}{16}=3\)
a, =\(\frac{\left(3\cdot4\right)^5}{4^3\cdot3^4}\)= \(\frac{3^4\cdot3\cdot4^3\cdot4^2}{4^3\cdot3^4}\)= 3 * 4^2 = 3 * 16 = 48
b, (310*11+38*45)/16*39=3^10*11+3^10*5/16*3^9=3^10*16/3^9*16=3
a, 12^5/64*3^4=2^10*3^5/2^6*3^4=2^4*3=48
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
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 \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)
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:
\(D+3D=\left(3-3^2+...+3^{11}\right)+\left(3^2-3^3+...+3^{12}\right)\)
\(\Leftrightarrow4D=3^{12}+3\)
\(\Rightarrow D=\frac{3^{12}+3}{4}\)