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\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(A=1-\dfrac{1}{n^2+2n+1}\)
\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(=1-\dfrac{1}{n^2+2n+1}\)
\(=\dfrac{n^2+2n}{n^2+2n+1}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
\(1,\)
\(a,\) Sửa: \(A=10^n+72n-1⋮81\)
Với \(n=1\Leftrightarrow A=10+72-1=81⋮81\)
Giả sử \(n=k\Leftrightarrow A=10^k+72k-1⋮81\)
Với \(n=k+1\Leftrightarrow A=10^{k+1}+72\left(k+1\right)-1\)
\(A=10^k\cdot10+72k+72-1\\ A=10\left(10^k+72k-1\right)-648k+81\\ A=10\left(10^k+72k-1\right)-81\left(8k-1\right)\)
Ta có \(10^k+72k-1⋮81;81\left(8k-1\right)⋮81\)
Theo pp quy nạp
\(\Rightarrow A⋮81\)
\(b,B=2002^n-138n-1⋮207\)
Với \(n=1\Leftrightarrow B=2002-138-1=1863⋮207\)
Giả sử \(n=k\Leftrightarrow B=2002^k-138k-1⋮207\)
Với \(n=k+1\Leftrightarrow B=2002^{k+1}-138\left(k+1\right)-1\)
\(B=2002\cdot2002^k-138k-138-1\\ B=2002\left(2002^k-138k-1\right)+276138k+1863\\ B=2002\left(2002^k-138k-1\right)+207\left(1334k+1\right)\)
Vì \(2002^k-138k-1⋮207;207\left(1334k+1\right)⋮207\)
Nên theo pp quy nạp \(B⋮207,\forall n\)
\(2,\)
\(a,\) Sửa đề: CMR: \(1\cdot2+2\cdot3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
Đặt \(S_n=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
Với \(n=1\Leftrightarrow S_1=1\cdot2=\dfrac{1\cdot2\cdot3}{3}=2\)
Giả sử \(n=k\Leftrightarrow S_k=1\cdot2+2\cdot3+...+k\left(k+1\right)=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}\)
Với \(n=k+1\)
Cần cm \(S_{k+1}=1\cdot2+2\cdot3+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)
Thật vậy, ta có:
\(\Leftrightarrow S_{k+1}=S_k+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)
Theo pp quy nạp ta có đpcm
\(b,\) Với \(n=0\Leftrightarrow0^3=\left[\dfrac{0\left(0+1\right)}{2}\right]^2=0\)
Giả sử \(n=k\Leftrightarrow1^3+2^3+...+k^3=\left[\dfrac{k\left(k+1\right)}{2}\right]^2\)
Với \(n=k+1\)
Cần cm \(1^3+2^3+...+k^3+\left(k+1\right)^3=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)
Thật vậy, ta có
\(1^3+2^3+...+k^3+\left(k+1\right)^3\\ =\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3\\ =\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\\ =\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)
Theo pp quy nạp ta được đpcm
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
A= \(\dfrac{1^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
= \(\dfrac{1}{1\cdot3}\cdot\dfrac{3^2}{3\cdot5}\cdot\dfrac{5^2}{5\cdot7}\cdot...\cdot\dfrac{n^2}{n\left(n+2\right)}\)
=\(\dfrac{1}{n+2}\)
B = \(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}\)
= \(\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}=\dfrac{16}{1-x^{16}}\)
1: =>3x+1=4
=>3x=3
hay x=1
2: \(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^9}{98^3}=\dfrac{1}{2^3}+\dfrac{7^9}{7^6\cdot2^3}\)
\(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^3}{2^3}=\dfrac{344}{2^3}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
3: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\x-\dfrac{2}{9}=-\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{9}\end{matrix}\right.\)
4: =>x+2=0 và y-1/10=0
=>x=-2 và y=1/10
