Rút gọn biểu thức sau
S = 1 + 1/3 + 1/3^2 + 1/3^3 + .......+ 1/3^n
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LQ
1
T
25 tháng 12 2023
\(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(=\dfrac{\sqrt{3}-1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-2\sqrt{3}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}-2\sqrt{3}\)
\(=\dfrac{2\sqrt{3}}{2}-2\sqrt{3}\)
\(=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
BM
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
NL
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^n}\)
=>\(3S=3.\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}\right)=3+1+\frac{1}{3}+...+\frac{1}{3^{n-1}}\)
=>\(3S-S=\left(3+1+\frac{1}{3}+.....+\frac{1}{3^{n-1}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}\right)\)
=>\(2S=3+1+\frac{1}{3}+....+\frac{1}{3^{n-1}}-1-\frac{1}{3}-\frac{1}{3^2}-....-\frac{1}{3^n}=3-\frac{1}{3^n}=\frac{3^{n+1}-1}{3^n}\)
=>\(S=\frac{3^{n+1}-1}{3^n}:2=\frac{3^{n+1}-1}{3^n.2}\)
Vậy.................