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Ta có :
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{4-1}{1^2.2^2}+\frac{9-4}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Chúc bạn học tốt ~
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(3.2\right)^2}+...+\frac{2n+1}{\left[n.\left(n+1\right)\right]^2}\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{2n+1}{n^2.\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
mk chỉ làm được đến đấy thôi
a. \(\left(\frac{-1}{5}\right)^n=\frac{-1}{125}\)
<=> \(\left(\frac{-1}{5}\right)^n=\left(\frac{-1}{5}\right)^3\)
<=> n = 3
b. \(\left(\frac{-2}{11}\right)^m=\frac{4}{121}\)
<=> \(\left(\frac{-2}{11}\right)^m=\left(\frac{2}{11}\right)^2\)
<=> m = 2
c. 72n + 72n+2 = 2450
<=> 72n + 72n . 72 = 2450
<=> 72n.(1+72) = 2450
<=> 72n = 72
<=> 2n = 2
<=> n = 1
\(\dfrac{n-2}{4}=\dfrac{1}{4}\\ \Leftrightarrow\left(n-2\right).4=1.4\\ \Rightarrow4n-8=4\\ \Rightarrow4n=4+8\\ \Rightarrow4n=12\\ \Rightarrow n=12:4=3\)
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=2\)
a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow2^{-\left(2n-1\right)}=2^{-3}\)
\(\Rightarrow2^{-2n+1}=2^{-3}\)
\(\Rightarrow-2n+1=-3\)
\(\Rightarrow-2n=-4\)
\(\Rightarrow n=-2\)
Vậy ...
b) \(\left(\dfrac{7}{5}\right)^n=\dfrac{343}{125}\)
\(\Rightarrow\left(\dfrac{7}{5}\right)^n=\left(\dfrac{7}{5}\right)^3\)
\(\Rightarrow n=3\)
Vậy ....
a) \(5^{n+3}-5^{n+1}=5^{12}.120\Leftrightarrow5^{n+1}.\left(5^2-1\right)=5^{12}.5.24\)
\(\Leftrightarrow24.5^{n+1}=5^{13}.24\Leftrightarrow5^{n+1}=5^{13}\Leftrightarrow n+1=13\Leftrightarrow n=12\)
b) \(2^{n+1}+4.2^n=3.2^7\)
\(\Leftrightarrow2^n\left(2+4\right)=3.2^7\Leftrightarrow6.2^n=3.2^7\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)
c) \(3^{n+2}-3^{n+1}=486\)
\(\Leftrightarrow3^{n+1}.\left(3-1\right)=486\Leftrightarrow2.3^{n+1}=486\Leftrightarrow3^{n+1}=243\)
\(\Leftrightarrow3^n=243:3=81=3^3\Leftrightarrow n=3\)
d) \(3^{2n+3}-3^{2n+2}=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.\left(3-1\right)=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.2=2.3^{10}\Leftrightarrow3^{2n+2}=3^{10}\Leftrightarrow2n+2=10\Leftrightarrow2n=8\Leftrightarrow n=4\)
a, 5n+1 - 5n-1 = 1254.23.3
5n-1.(52 - 1) = 1254.24
5n-1.24 = 1254.24
5n-1 = 1254
5n-1 = (53)4
5n-1 = 512
n - 1 = 12
n = 12 + 1
n = 13
b,22n-1 + 22n+2 = 3.211
22n-1.(1 + 23) = 3.211
22n-1.9 = 3.211
22n-1 = 211: 3
22n = 212 : 3 (xem lại đề bài em nhá)