Phú Nguyễn bạn sai rồi

lay so cuoi cong so dau roi chia 2 nhe

Điên à, bài này mà làm thế thì quay về lớp 1 học đi

(4/ 3* 9/ 8* 16/ 15* 25/ 24......9801/ 9800)

=(2* 2* 3* 3* 4* 4* 5* 5......*99* 99)/ (1* 3* 2* 4* 3* 5* 4* 6......98* 100)

=((2* 3* 4* 5*......*99)/ (1* 2* 3* 4*....* 98))/ ((2* 3* 4*...99)/ (3* 4* 5*.....*100))

=99* (1/ 50)

=99/ 50

đúng thì k cho mình bn nha (^.^)

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\)

\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)

\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)

\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)

.......

\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)

\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)

\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)

= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)

= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)

= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)

\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)

Ta thấy khoảng cách của các số lần lượt là : 

8 ; 16 ; 24 và chúng đều chia hết cho 8

Còn lại  tự làm nhé

:))

Ta có 1+9+25+49+...+9801

=1²+3²+5²+7²+...+99²

Ta có công thức tổng quát

1²+3²+5²+...+(2n-1)²=\(\frac{n\left(4n^2-1\right)}{3}\)

Ta có 99=2x50-1

=>1²+3²+5²+...+99²=\(\frac{50.\left(4.50^2-1\right)}{3}=166650\)

Cộng 2 vế với:4+16+36+...+10000

   1+9+25+49+...+9801+(4+16+36+...+10000

=1+4+9+16+25+...+9801+10000

=1x1+2x2+3x3+4x4+5x5+...100x100

=1x(2-1)+2x(3-1)+3x(4-1)+...+100x(101-1)

=1x2-1+2x3-2+3x4-3+...+100x101-100

=(1x2+2x3+...+100x101)-(1+2+3+...+100)

=(100x101x102):3-(101x100:2)

=343400-5050

=338350

cho mik nhé

Từ đề bài suy ra: A=3/4 x 8/9 x ...x 9800/9801 x 9999/10000

                       =>A=<1x3/2x2> x <2x4/3x3> x ... x <99x101/100x100>

                       =>A=(1x2x...x99)/(2x3x...x100) x (3x4x...x101)/(2x3x...x100) 

                       =>A=1/100 x 101/2 = 101/200

BẠn hãy tính ra rồi phân tích tử và mẫu là ra

so tiep theo la96059601

quy luat; số lien sau bằng số truoc bình phương