Đặt \(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(3A=999900\)

\(A=999900:3\)

\(A=333300\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}+\frac{1}{15.16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

Chúc bạn học giỏi nha!!!

K cho mik vs nhé 

1/1.2+1/2.3+1/3.4+.........+1/14.15+1/15.16

=1-1/2+1/2-1/3+...+1/15-1/16

=1-1/16

=15/16  *k mk nha*

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}+\frac{1}{15.16}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{16}{16}-\frac{1}{16}=\frac{15}{16}\)

3S= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)] 
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)] 
=n(n+1)(n+2) 
=>S 

Nguyễn Đình Phương giống cn gái tek?

Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101  3S = 3.33.100.101 
 S=33.100.101= 333300

số các chữ số đó là

(200-1):1+1=200

số cặp đó là

200:2=100

tổng 1 cạp là

200+1=201

giá trị bt là 

201.100=20100

Ta có : \(S=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{1024}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+.....+\frac{1}{1024}\)

\(\Rightarrow2S-S=2-\frac{1}{1024}\)

\(\Rightarrow S=\frac{2047}{1024}\)

s=1.2+2.3+3.4+...+99.100

=>3s=1.2.3+2.3.3+3.4.3+...+99.100.3

=1.2.3+2.3.(4-1)+...+99.100.(101-98)

=1.2.3-1.2.3+2.3.4-2.3.4+...-98.99.100+99.100.101

=99.100.101

=>s=99.100.101/3=333300

\(\text{ta có: }\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

             \(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

           \(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)

           ...........................

             \(\frac{1}{39.40}=\frac{1}{39}-\frac{1}{40}\)

Đồng nhất 2 vế ta có:

   \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{39}-\frac{1}{40}=\frac{1}{2}-\frac{1}{40}=\frac{19}{40}\)

Ta có :

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)

\(=\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)

\(=\)\(\frac{1}{2}-\frac{1}{40}\)

\(=\)\(\frac{20}{40}-\frac{1}{40}\)

\(=\)\(\frac{19}{40}\)

Vậy \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}=\frac{19}{40}\)