Ảnh Phneas and Ferb đẹp đấy, hơn cả của mình !

ủa , vậy qui tắt của dãy số này là gì ?

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{5050}=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{10100}\right)\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\right)=2.\frac{99}{202}=\frac{99}{101}\)

Đặt A = 1/3+1/6+1/10+1/15+...+1/5050

A : 2 ta có : 1/6+1/12+1/20+1/30+...+1/10100

A: 2 = 1/2x3+1/3x4+1/4x5+1/5x6+..... + 1/ 100 x 101

A: 2 = 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...1/100-1/101

Rút gọn ta được :

 A: 2 = 1/2-1/101

A: 2 = 99/202

A = 99/202x2 = 99 / 101

\(C=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+\frac{3}{10}+....+\frac{3}{5050}\)

\(C=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+\frac{6}{20}+...+\frac{6}{10100}\)

\(C=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+\frac{6}{4.5}+....+\frac{6}{100.101}\)

\(C=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(C=6.\left(1-\frac{1}{101}\right)=6.\frac{100}{101}=\frac{600}{601}\)

Vậy \(C=\frac{600}{601}\)

\(D=\frac{1}{1.3.5}+\frac{1}{3.5.7}+....+\frac{1}{2009.2011.2013}\)

\( D=\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{2009.2011.2013}\right)-\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{2009.2011.2013}\right)\)

\(D=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{2009.2011}-\frac{1}{2011.2013}\right)\)

\(D=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{2011.2013}\right)=\frac{1}{4}.\frac{1349380}{4048143}=\frac{1349380}{16192572}\)

Vậy \(D=\frac{1349380}{16192572}\)

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