Ta có:\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{210}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{14.15}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{14}-\frac{1}{15}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)

\(=1-\frac{2}{15}=\frac{15-2}{15}=\frac{13}{15}\)

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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+...+\(\dfrac{1}{182}\)+ 210

 A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{13\times14}\)+ 210

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+\(\dfrac{1}{13}\) - \(\dfrac{1}{14}\) + 210

A = 1 - \(\dfrac{1}{14}\) + 210

A = 211 - \(\dfrac{1}{14}\)

A = \(\dfrac{2953}{14}\)

Dãy số cách đều: 2;4;6;8...; b

Có số số hạng là: \(\frac{\left(b-2\right)}{2}+1=\frac{b}{2}\)

Có tổng là: \(\frac{1}{2}\cdot\left(2+b\right)\cdot\frac{b}{2}=210\Leftrightarrow b\left(b+2\right)=840=28\cdot30=28\cdot\left(28+2\right)\)

Do đó, b = 28

a) 39                    b) 20                        c) 29                       d) ...

S=2/6+2/12+2/20+...+2/90

S=2/2.3+2/3.4+...+2/9.10

S=2(1/2.3+...+1/9.10)

S=2(1/2-1/3+...+1/9-1/10)

S=2.2/5

S=4/5

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{90}\)

\(=\frac{2}{2x3}+\frac{2}{3x4}+\frac{2}{4x5}+...+\frac{2}{9x10}\)

\(=\left(\frac{2}{2x3}+\frac{2}{3x4}+\frac{2}{4x5}+...+\frac{2}{9x10}\right):2x2\)

\(=\left(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\right)x2\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)x2\)

\(=\left(\frac{1}{2}-\frac{1}{10}\right)x2\)

\(=\frac{4}{5}x2\)

\(=\frac{8}{5}\)