\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(=1-\dfrac{1}{n^2+2n+1}\)

\(=\dfrac{n^2+2n}{n^2+2n+1}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Xét thừa số tổng quát:

\(\dfrac{k}{\left(\dfrac{k-1}{2}.\dfrac{k+1}{2}\right)^2}\)\(=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}\)

\(=\dfrac{k}{\dfrac{\left[\left(k-1\right)\left(k+1\right)\right]^2}{16}}=\dfrac{k}{\dfrac{\left(k^2-1\right)^2}{16}}=\dfrac{16k}{\left(k^2-1\right)^2}\)

Thay \(k=3;5;....2n+1\) ta được:

\(\dfrac{16.3}{\left(3^2-1\right)^2}+\dfrac{16.5}{\left(5^2-1\right)^2}+....+\dfrac{16.n}{\left(n^2-1\right)^2}\)

\(=16.\left(\dfrac{3}{\left(3^2-1\right)^2}+\dfrac{5}{\left(5^2-1\right)^2}+...+\dfrac{n}{\left(n^2-1\right)^2}\right)\)

\(=16.\left(\dfrac{3}{\left[\left(3-1\right)\left(3+1\right)\right]^2}+\dfrac{5}{\left[\left(5-1\right)\left(5+1\right)\right]^2}+...+\dfrac{n}{\left[\left(n-1\right)\left(n+1\right)\right]^2}\right)\)

\(=16.\left(\dfrac{3}{4.16}+\dfrac{5}{16.36}+...+\dfrac{n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{12}{4.16}+\dfrac{20}{16.36}+...+\dfrac{4n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{36}+...+\dfrac{1}{\left(n-1\right)^2}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2}{4\left(n+1\right)^2}-\dfrac{4}{4\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2-4}{4\left(n+1\right)^2}\right)=\dfrac{4\left(n+1\right)^2-16}{4\left(n+1\right)^2}\)

\(=\dfrac{4\left[\left(n+1\right)^2-4\right]}{4\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-4}{\left(n+1\right)^2}\)

Chúc bạn học tốt!!!

Lời giải:

Xét số hạng tổng quát:

\(\frac{2n+1}{[n(n+1)]^2}=\frac{1}{n(n+1)}.\frac{2n+1}{n(n+1)}=\frac{n+1-n}{n(n+1)}.\frac{n+(n+1)}{n(n+1)}\)

\(=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n}+\frac{1}{n+1}\right)=\frac{1}{n^2}-\frac{1}{(n+1)^2}\)

Do đó:

\(S=\frac{3}{(1.2)^2}+\frac{5}{(2.3)^2}+....+\frac{2n+1}{[n(n+1)]^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{(n+1)^2}\)

\(=1-\frac{1}{(n+1)^2}\)

Ta có:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)

\(=1-\frac{2n+1}{\left(n+1\right)^2}\)

Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)

SAI RỒI ĐÁP ÁN LÀ N^2/(N+1)^2

Chọn B

\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)

\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)

\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)

\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)

\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)

a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)

\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)

\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)

\(x=\dfrac{-9198}{4400}\)

a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)

\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(x+\dfrac{206}{100}=5\)

\(x=5-\dfrac{206}{100}\)

\(x=\dfrac{147}{50}\)

Vậy \(x=\dfrac{147}{50}\)

\(1,\)

\(a,\) 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