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Đặt \(A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\)

\(=\frac{3}{1\times2}+\frac{3}{2\times3}+\frac{3}{3\times4}+...+\frac{3}{99\times100}\)

\(\Rightarrow A:3=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

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

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

\(\Rightarrow A=\frac{99}{100}\times3=\frac{297}{100}\)

Vậy \(A=\frac{297}{100}\).

mình chịu lun .xin lỗi!

Kocho Shinobu

\(=3.\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)

\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

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

\(=3.\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}=\frac{297}{100}\)

\(\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\\ =3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

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

\(=3\left(1-\frac{1}{100}\right)=3.\frac{99}{100}=\frac{297}{100}\)

\(\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{9900}\)

\(=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{99.100}\)

\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=3.\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}\)

\(=\frac{297}{100}\)

Tui ko bít 

Giải 

\(A=1+2+3+4+5+...+99+100\)

Số số hạng của A là: \(\left(100-1\right)\div1+1=100\)(số hạng)

Tổng A là: \(\frac{\left(100+1\right)\times100}{2}=5050\)

Vây A=5050

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

\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

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

\(B=1-\frac{1}{100}=\frac{99}{100}\)

Vậy \(B=\frac{99}{100}\)

minh cam thay de hoi sai

A = 1 + 2 + 3 + ... + 99 + 100

A = 100 + 99 + ... + 2 + 1

2A = 101 + 101 +... + 101 + 101 ( 100 số hạng )

A = 101 . 100 : 2 = 5050

Vậy A = 5050 

B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/9900

B = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100

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

B = 1/1 - 1/100

B = 99/100

Vậy B = 99/100

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

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

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

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

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

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

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

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

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

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

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