b=1²+2²+3²+............+98²

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

b=1.2-1+2.3-2+3.4-3+.............+98.99-98

b=(1.2+2.3+3.4+..................+98.99)-(1+2+3+............+98)

a-b=1+2+3+...............+98

a-b=\(\frac{98.\left(98+1\right)}{2}\)

a-b=4851

                                   Vậy A-B=4851

A=1-1/2+1/2-1/3+1/3-1/4

A=1-1/4=3/4

ai k mk mk k lại

A = 1 - 1/2 + 1/2 - 1/3 - 1/4

A = 1 - 1/4 = 3/4

1.2+2.3+1+1+1=11

tk minh di

= 2 + 6 + 3

= 8 + 3

= 11

??? Đăng cái j z

ủa toán lớp mấy chứ ko phải lớp 1

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\frac{50}{201}\)

\(A=\frac{25}{101}\)

=25/101

k cho to nhe

Cái này mà Toán lớp 1 tui xỉu.

Mà mk làm xong h cho mk nha.

ko chep de

= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1/2 - 1/100

= 49/100

=)) tích nha : j

1/2*3+1/3*4+.....+1/99*100

=1/2-1/3+1/3-1/4+....+1/99-1/100

=1/2-1/100

=50/100-1/100

=49/100

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 1\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

A=1-\(\frac{1}{10}\)

A=\(\frac{9}{10}\)

\(\frac{1}{1.2}.\frac{1}{2.3}....\frac{1}{9.10}=\frac{1.1.1.1.1.1}{1.2.2.3.3....9.9.10}=\frac{1}{1.4.9.16.25.36....100}=\frac{1}{13168189440000}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)