\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)

bạn k trước mk mới kb

=1/125

bằng4050

4150

1+2+3+...+99+100

=100*101/2

=5050

cho 1+5+5^2+5^3+...+5^100 =A

ta có A = 1+5+5^2+5^3+...+5^100

=>5A =5 +5^2+5^3+5^4+...+5^101

=>5A-A=(5+5^2+5^3+...+5^101)-(1+5+5^2+5^3+...+5^100)

=>4A=5^101-1

=>A =(5^101-1)/4

vậy 1+5+5^2+5^3+...+5^100=(5^101-1)/4
 

5A=5+5^2+5^3+...+5^100+5^101

5A-A=[5+5^2+...+5^101]-[1+5+5^2+...+5^100]

4A=1+5^101

VẬY A=\(\frac{1+5^{101}}{4}\)NOTE ĐÚNG NHA

tìm B khi : B=5^2/10x15+5^2/15x20+...+5^2/190x195+5^2/195x200

=(100-55)+(99-54)+(98-53)+...+(46-1) (có 55 tổng)

=45 x 55=2475

2475 nha

\(S=1\times2+2\times3+3\times4+...+99\times100\)

\(3\times S=1\times2\times3+2\times3\times\left(4-1\right)+3\times4\times\left(5-2\right)+...+99\times100\times\left(101-98\right)\)

\(=1\times2\times3+2\times3\times4-1\times2\times3+3\times4\times5-2\times3\times4+...+99\times100\times101-98\times99\times100\)

\(=99\times100\times101\)

\(S=\frac{99\times100\times101}{3}\)

1/5051 +(-2/5051)+3/5051+(-4/5051)+...+99/5051+(-100/5051)

\(=\frac{\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)}{5051}\)

\(=\frac{-1+\left(-1\right)+..+\left(-1\right)}{5051}\) (có 50 số -1)

\(=\frac{\left(-1\right).50}{5051}=\frac{-50}{5051}\)

1 - 2 + 3 - 4 + 5 - 6 +...+ 99 - 100

= ( 1- 2 ) + ( 3 - 4 ) + ( 5 - 6 ) +....+ ( 99- 100)

= 1 + 1 + 1 +....+1 ( 50 c/s 1 )

= 1. 50 

=50

-50 nha bn

1-2 + 3-4 + 5-6 + ... + 99-100

=> (1-2) + (3-4) + (5-6) + ... + (99-100) . Ta có : 50 cặp như vậy

=> (-1) + (-1) + (-1) + ... + (-1)

=> (-1) . 50

= -50