b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=3.\frac{99}{100}\)

\(\Rightarrow A=\frac{297}{100}\)

\(A=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{97.100}\)

\(A=9\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=9.\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...-\frac{1}{100}\right)\)

\(A=\frac{9}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=3\left(\frac{99}{100}\right)=\frac{297}{100}\)

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=3\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}\)

\(A=\dfrac{297}{100}\)

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)

\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)

\(A=\frac{3}{1}-\frac{3}{100}\)

\(A=\frac{297}{100}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(\Rightarrow A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)

\(A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

= 3/1 - 3/4 + 3/4 - 3/7 + 3/7 - 3/10 + ... + 3/97 - 3/100 = 3/1 - 3/100 = 297/100

\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)

\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=3.\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)

\(A=\dfrac{3^2}{1\times4}+\dfrac{3^2}{4\times7}+\dfrac{3^2}{7\times10}+\dfrac{3^2}{10\times13}+\dfrac{3^2}{13\times16}...+\dfrac{3^2}{97\times100}\)

\(=3\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16} +...+\dfrac{3}{97\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

\(=2\dfrac{97}{100}\)

Vậy \(A=2\dfrac{97}{100}\)