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

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

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

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

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

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

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

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

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

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

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

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

\(=1386010912\)

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

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

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

=2765871900

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

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

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

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

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

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

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

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

\(=2049300^2\)

(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) =     1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)

(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1  - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))

\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))

   \(\dfrac{1}{2022}\).\(x\)        = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))

   \(\dfrac{1}{2022}\).\(x\)        =    \(\dfrac{1}{2003}\)

             \(x\)        = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)

             \(x\)       =     \(\dfrac{2022}{2003}\)

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

(1-1/2).(1-1/3).(1-1/4)...(1-1/2002).x=1-1/1.2-1/2.3-1/3.4-...-1/2002.2003 ghi loi giai nha ae

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)

C=1*2+2*3+3*4+...+98*99

C=2+6+12+...+9702

C=2+9702

C=9704

vay C=9704

D=(1*99+2*99+3*99+...+99*99)-(1*2+2*3+3*4+...+98*99)

D=(99+198+297+...+9801)-(2+6+12+...+9702)

D=(99+9801)-(2+9702)

D=9900-9704

D=196

vay D=196

ai di qua dong tinh thi nho h cho minh nhe

=43.61

Lời giải:

$A=1.2+2.3+3.4+...+50.51$
$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+...+50.51(52-49)$

$=(1.2.3+2.3.4+3.4.5+...+50.51.52)-(0.1.2+1.2.3+2.3.4+....+49.50.51)$

$=50.51.52$

$\Rightarrow A=50.51.52:3=44200$