Các số có tổng từ 1->100 có tổng là:2600

Có 200 số 2 nên ta lấy

2600.200=520 000

=>D=520 000

bạn ơi làm rõ hơn đi mình ko hiểu

\(A=1\cdot2^2+2\cdot3^2+\cdots+2017\cdot2018^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+2018^2\left(2018-1\right)\)

\(=\left(2^3+3^3+\cdots+2018^3\right)-\left(2^2+3^2+\cdots+2018^2\right)\)

\(=\left(1^3+2^3+3^3+\cdots+2018^3\right)-\left(1^2+2^2+3^2+\cdots+2018^2\right)\)

\(=\left(1+2+\cdots+2018\right)^2-\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}\)

\(=\left(2018\cdot\frac{2019}{2}\right)^2-\frac{2018\cdot2019\cdot4037}{6}=\left(1009\cdot2019\right)^2-1009\cdot673\cdot4037\)

=2.2.5²-2.3²

=2.50-2.9

=2.(50-9)

=2.41=82

Q=1.2.(3-1)+2.3.(4-1)+3.4.(5-1)+...+19.20.(21-1)=

=(1.2.3+2.3.4+3.4.5+...+19.20.21)-(1.2+2.3+3.4+...+19.20)

Đặt

A=1.2.3+2.3.4+3.4.5+...+19.20.21

4A=1.2.3.4+2.3.4.4+3.4.5.4+...+19.20.21.4=

=1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+19.20.21.(22-18)=

=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-18.19.20.21+19.20.21.22=

=19.20.21.22

\(A=\dfrac{19.20.21.22}{4}=5.19.21.22\)

Đặt 

B=1.2+2.3+3.4+...+19.20

3B=1.2.3+2.3.3+3.4.3+...+19.20.3=

=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)=

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-18.19.20+19.20.21=

=19.20.21

\(B=\dfrac{19.20.21}{3}=7.19.20\)

Q=A-B

câu hai thiếu đề bạn ạ

ko tính dc bạn ơi

A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

Ảnh hiển thị bị lỗi hết rồi bạn. Bạn coi lại.

hvt