\(T=3+8+15+24+35+...+9800+9999\) (sửa 3+5 thành 3+8)

Ta thấy :

\(3=2^2-1\)

\(8=3^2-1\)

\(15=4^2-1\)

\(24=5^2-1\)

\(.....\)

\(9800=99^2-1\)

\(9999=100^2-1\)

\(\Rightarrow T=1^2+2^2+3^2+...+100^2+\left(-1\right).100\)

\(\Rightarrow T=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-100\)

\(\Rightarrow T=\dfrac{100.101.201}{6}-100=338350-100=338250\)

sao phải sửa 5 thành 8 thế

B = \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\) + ..... + \(\frac{2}{99.101}\)+ \(\frac{2}{101.103}\)

 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/99 - 1/101 + 1/101 - 1/103

= 1- 1/103 = 102/103

(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 (^.^)

\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{9800}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{9801}{9800}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{99.99}{98.100}\)

\(=\frac{2.3.4...99}{1.2.3....98}.\frac{2.3.4...99}{3.4.5...100}\)

\(=99.\frac{2}{100}=99.\frac{1}{50}=\frac{99}{50}\)

aweee tks <3

ko có j

Ta có: \(A=\dfrac{5}{8}+\dfrac{5}{24}+\dfrac{5}{48}+...+\dfrac{5}{9800}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+...+\dfrac{2}{9800}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{49}{50}\)

\(=\dfrac{245}{100}=\dfrac{49}{20}\)

\(B=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right)...\left(99.101\right)}{2^2.3^2.4^2.5^2...100^2}=\frac{\left(1.2.3.4...99\right).\left(3.4.5.6...101\right)}{\left(2.3.4.5...100\right)\left(2.3.4.5...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)

B = \(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}\frac{4.6}{5^2}...\frac{99.101}{100^2}=\frac{1.3.2.4.3.5.4.6...99.101}{2.2.3.3.4.4.5.5...100.100}\)

   =\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

Vật B = \(\frac{101}{200}\)

đúng cái đi