Ta có :

\(D=1.1!+2.2!+...+100.100!\)

\(=\left(2-1\right)1!+\left(3-1\right).2!+\left(4-1\right).3!+...+\left(101-1\right).100!\)

\(=2!-1!+3!-2!+4!-3!+...+101!-100!\)

\(=101!-1!\)

Số quá lớn nhé :)

A=1x2+2x3+3x4+...+49x50

3A= 3(1.2+2.3+3.4+...+49.50)

3A= 1.2.3+2.3.3+3.4.3+...+49.50.3

3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)

3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51

3A= 49.50.51

A= 49.50.51/3=41650

B=1x3+3x5+5x7+...+99x101

B=1/1.3 +1/3.5 +...+1/99.101

2B=2/1.3 + 2/3.5 +...+2/99.101

2B=1-1/3+1/3-1/5+...+1/99-1/101

2B=1-1/101

2B=100/101

B=100/101:2=100/202

C=1x2x3+2x3x4+3x4x5+...+50x51x52
Nhân C với 4 ta được:
C x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+50x51x52x4
C x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 50x51x52x(53-49)
C x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... +49x 50x51x52 - 50x51x52x53
Sau khi cộng - trừ giản ước ta có : C x 4 = 50x51x52x53
C = 50x51x52x53 : 4 = 1756950

2.2+3.3+4.4+...+100.100

= 2²+3²+4²+...+100²

= 1²+2²+3²+...+100²-1

=\(\dfrac{100.\left(100+1\right).\left(2.100+1\right)}{6}\)-1

=338350-1

=338349

A= \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}\)

=> A= \(\frac{99}{100}>\frac{25}{26}\)

A=1+1+1+...+1

A=100x1

A=100