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31 tháng 1
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
TT
3
DP
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)
\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)
\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)
\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(M=11\cdot\left(1-\frac{1}{103}\right)\)
\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)
MC
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3.\frac{63}{268}\)
\(\Leftrightarrow Q=\frac{189}{268}\)
Câu b) bạn làm tương tự nhé :)
6 tháng 2 2022
\(S=4\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=4\cdot\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4\cdot\dfrac{32}{99}=\dfrac{128}{99}\)
TB
2
H
26 tháng 2 2023
\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)
26 tháng 2 2023
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)
15 tháng 7 2018
\(S_1=\frac{5}{20\times22}+\frac{5}{22\times24}+...+\frac{5}{88\times90}\)
\(S_1=\frac{5}{2}\times\left(\frac{2}{20\times22}+\frac{2}{22\times24}+...+\frac{2}{88\times90}\right)\)
\(S_1=\frac{5}{2}\times\left(\frac{1}{20}-\frac{1}{22}+\frac{1}{22}-\frac{1}{24}+...+\frac{1}{88}-\frac{1}{90}\right)\)
\(S_1=\frac{5}{2}\times\left(\frac{1}{20}-\frac{1}{90}\right)\)
\(S_1=\frac{5}{2}\times\frac{7}{180}\)
\(S_1=\frac{7}{72}\)
_Chúc bạn học tốt_
HD
1
TT
14 tháng 4 2016
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
\(\frac{2^2}{1.3}+\frac{2^2}{3.5}+\frac{2^2}{5.7}+...+\frac{2^2}{99.101}\)
=2.\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
=2.\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}\right)\)
=2.\(\left(1-\frac{1}{100}\right)\)
=2.\(\left(\frac{100}{100}-\frac{1}{100}\right)\)
=2.\(\frac{99}{100}\)
=\(\frac{99}{50}\)