5P=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100

5P=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+....+97.98.99.100.101-96.97.98.99.100

5P=97.98.99.100.101

5P=9505049400

S=1901009880

P = 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + 4.5.6.7 + .. + 97.98.99.100

4P = ( 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + .. + 98.99.100) 4

4P = 1.2.3.(4-0) + 2.3.4(5-1) + 3.4.5(6-2) + 4.5.6(7-3) + 98.99.100(101-97)

4P = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + 4.5.6.7 - 3.4.5.6 + .. 98.99.100.101 - 97.98.99.100

4P = 98.99.100.101

4P= 98.99.100.101/4

Nếu thấy đúng thì tích mk nha

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)

=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)

=> \(A=\frac{451}{8120}\)

Các bạn giải giùm mình nhanh nhanh được k? Mình đang cần gấp

\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\\ A=\frac{1}{75}\)

\(B=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146+150}=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{94}+\frac{15}{94}-\frac{15}{98}+...+\frac{15}{146}-\frac{15}{150}\right)\)

\(B=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{150}\right)=\frac{1}{60}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

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

\(2A=1-\frac{1}{3^{100}}\)

\(A=\frac{1-\frac{1}{3^{100}}}{2}\)

\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)

\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)

\(B=\frac{15}{14}:3=\frac{5}{14}\)

a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

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

\(2A=1-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)

b)  \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\frac{3}{14}\)

\(\Rightarrow B=\frac{5}{14}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=1-\frac{1}{3^{100}}\)\(\Rightarrow2A=\left(1-\frac{1}{3^{100}}\right)\Rightarrow A=\frac{1}{2}\times\left(1-\frac{1}{3^{100}}\right)\)

a) F = \(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)+\frac{1}{2}.\left(\frac{1}{27}-\frac{1}{29}\right)+\frac{1}{2}.\left(\frac{1}{29}-\frac{1}{31}\right)+...+\frac{1}{2}.\left(\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\frac{2}{75}\)

F = \(\frac{1}{75}\)

b) G = \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)

G = \(\frac{15}{4}.\frac{4}{90.94}+\frac{15}{4}.\frac{4}{94.98}+\frac{15}{4}.\frac{4}{98.102}+...+\frac{15}{4}.\frac{4}{146.150}\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}\right)+\frac{15}{4}.\left(\frac{1}{94}-\frac{1}{98}\right)+\frac{15}{4}.\left(\frac{1}{98}-\frac{1}{102}\right)+...+\frac{15}{4}.\left(\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\frac{1}{225}\)

G = \(\frac{1}{60}\)

<br class="Apple-interchange-newline"><div id="inner-editor"></div>12.4 +14.6 +...+198.100 

=12 (22.4 +24.6 +...+298.100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −1100 )=12 .49100 =49200 

1056 +10140 +10260 +...+101400 =53 (

c) H = \(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

H = \(\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

H = \(\frac{5}{3}.\frac{3}{28}+\frac{5}{3}.\frac{3}{70}+\frac{5}{3}.\frac{3}{130}+...+\frac{5}{3}.\frac{3}{700}\)

H = \(\frac{5}{3}.\left(\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+...+\frac{3}{700}\right)\)

H = \(\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\frac{3}{14}\)

H = \(\frac{5}{14}\)

Tk mình đi mọi người mình bị âm nè!

ai tk mình mình tk lại cho!!!

trả lời câu hỏi của mình đã

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