\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

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

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

Giải:

\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)

Đk: \(n\ne0;n\ne-1\)

\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)

\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)

\(\Leftrightarrow C=\dfrac{n+2}{3n}\)

Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)

\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)

Vậy ...

Giải:

\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)

Đk: \(n\ne0;n\ne-1\)

\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)

\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)

\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)

\(\Leftrightarrow C=\dfrac{n+2}{3n}\)

Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)

\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)

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

Bạn ơi (n+1)(n+2) hay (n+1)-(n+2) vậy

var n,i:integer;

s:real;

begin

write('n=');readln(n);

s:=0;

for i:=1 to n do s:=s+(1/(i*(i+1)));

writeln(' Tong la: ',s);

readln;

end.

2) var n,i:integer;

s:real;

begin

write('n=');readln(n);

s:=0;

for i:=1 to n do s:=s+(1/((2*i)-1));

writeln(' Tong la: ',);

readln;

end.

sửa đề: S = \(1+\dfrac{1}{2.3}+\dfrac{2}{3.4}+...+\dfrac{n}{\left(n+1\right).\left(n+2\right)}\)

var i,n:integer;

    s:real;

begin

write('Nhap n = ');readln(n);

s:=1;

for i:=1 to n do

s:=s+i/((i+1)*(i+2));

write('S = ',s:10:2);

readln;

end.

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

ngu như con bò tót, ko biết 1+1=2.