a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{100}-\frac{1}{101}\)

=\(1-\frac{1}{101}\)

=\(\frac{100}{101}\)

b,\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)

=\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{256}-\frac{1}{512}\right)\)

=\(1-\frac{1}{512}\)

=\(\frac{511}{512}\)

\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}+\frac{1}{10100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

a) đề sai

    1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100

= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101

= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101

= 1/1 - 1/101

= 100 /101

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

=\(1-\frac{1}{101}\)

=\(\frac{100}{101}\)

Mik trả lời ở bài dưới rồi đó.

o dau ha ho thu giang

Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)

     \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

     \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

     \(=1-\frac{1}{101}\)

     \(=\frac{100}{101}\)

Tương đương \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

A = 1+ 1+1+ ...+ 1 +(\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}+\dfrac{1}{10100}\))

=(1+1+1+...+1)+ (\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}+\dfrac{1}{100x101}\))

=100 +\(1-\dfrac{1}{101}=100-\dfrac{100}{101}=\dfrac{10000}{101}\)

1+1/2+1+1/6+1+1/12+...+1+1/9900

=1+1/1*2+1+1/2.3+....+1+1/99*100

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

=100+99/100

=10099/100

1/1x2+1/2x3+1/3x4+...+1/99x100+1/100x101

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100+1/100-1/101

=1/1-1/101

=100/101

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)

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

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

\(=\frac{4}{5}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

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

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

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

Từng ý một nhanh hơn nhá

Đặt A=1+2+2²+2³+...+2¹⁰⁰

suy ra 2A=2+2²+2³+...+2¹⁰⁰

suy ra 2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+2³+...+2¹⁰⁰⁾

                 =2¹⁰¹-1

Vậy 1+2+2²+2³+...+2¹⁰⁰=2¹⁰¹-1

 A=1+2+2^2+2^3+...+2^100

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

Vậy A= 2¹⁰¹-1

1) x=100