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\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

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

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

2 tháng 8 2015

bạn lên mạng tra từng câu 1 sẽ có

3 tháng 8 2015

ukm cảm ơn bạn nhìu

 

7 tháng 1 2016

Bài 3 : a) 3784 + 23 - 3785 - 15

= (3784 - 3785) + (23 - 15)

= -1 + 8 

= 7

b) 21 + 22 + 23 + 24 - 11 - 12 - 13 - 14

= (21 - 11) + (22 - 12) + (23 - 13) + (24 - 14)

= 10 + 10 + 10 + 10 

= 40

Bài 4 : a) -2001 + (1999 + 2001)

= -2001 + 1999 + 2001

= ( - 2001 + 2001 ) + 1999

= 0 + 1999

= 1999

B) (43 - 863) - (137 - 57)

= 43 - 863 - 137 - 57 

= (43 - 57) + ( -863 - 137 )

= -14 + -1000

= -1014 

Nhớ tick !!!

7 tháng 1 2016

bài 3 

a.=22

b.=40

bài 4:

a.=1999

b.=-900

26 tháng 4 2017

A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=1\(-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(\dfrac{47}{60}\)

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

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}+\dfrac{1}{101}\)

=\(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\)

25 tháng 4 2017

A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)

= \(\dfrac{47}{60}\)

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

= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

= 2\(\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{200}{101}\)

10 tháng 12 2016

Bài 1 : Ta có : a = 1.2 + 2.3 + 3.4 + ....... + 99.100

=> 3a = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...... + 99.100.(101 - 98)

=> 3a = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101

=> 3a = 99.100.101

=>   a = \(\frac{99.100.101}{3}=333300\) 

16 tháng 9 2018

vào phần câu hỏi tương tự là có ấy bạn

9 tháng 9 2017

a)Đặt \(A=\dfrac{6}{1.4}+\dfrac{6}{4.7}+\dfrac{6}{7.10}+...+\dfrac{6}{97.100}\)

\(3a=3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{97}-\dfrac{3}{100}\)

\(=3-\dfrac{3}{100}\)

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

b)Đặt \(B=\dfrac{4}{1.3}+\dfrac{16}{3.5}+\dfrac{36}{5.7}+...+\dfrac{9604}{97.99}\)

\(=2b=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

\(2b=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{97}-\dfrac{2}{99}\)

\(2b=2-\dfrac{2}{99}=\dfrac{198}{99}-\dfrac{2}{99}=\dfrac{196}{99}\)

c) Tương tự! Bạn tự làm nhé!

Bài 5:

a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)

\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)

\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)

hay A=330

Vậy: A=330