S = 1/100

Đặt A = 1×2 + 2×3 + 3×4 + ... + 20×21

3A = 1×2×3 + 2×3×(4-1) + 3×4×(5-2) + ... + 20×21×(22-19)

= 1×2×3 - 1×2×3 + 2×3×4 - 2×3×4 + 3×4×5 - ... - 19×20×21 + 20×21×22

= 20×21×22

A = 20×21×22 : 3

= 20×22×7

= 3080

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)

Lời giải:
Gọi biểu thức số 1 là A và số 2 là B

\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)

help me , pls

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3 A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) ..................................

A x 3 = 99x100x101 A = 333300

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
..................................
A x 3 = 99x100x101
A = 333300

A = \(\frac{1}{1\times2}\) + \(\frac{1}{2\times3}\) + \(\frac{1}{3\times4}\) + ... + \(\frac{1}{19\times20}\) + \(\frac{1}{20\times21}\)

A = \(\frac11\) - \(\frac12\) + \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + ... + \(\frac{1}{19}\) - \(\frac{1}{20}\) + \(\frac{1}{20}\) - \(\frac{1}{21}\)

A = \(\frac11\) - \(\frac{1}{21}\)

A = \(\frac{20}{21}\)

Ta có: \(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{20\times21}\)

\(=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{20}-\frac{1}{21}\)

\(=1-\frac{1}{21}=\frac{20}{21}\)