2/5+2/35+2/63+2/99+: +2/483
2/5+2/35+2/63+2/99+ +2/483 = ?
2/15+2/35+2/63+2/99+ 2/483 = ?
2/15+2/35+2/63+2/99+2/143
=2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
= 1/3 - 1/13
= 13/39 -3/39
=10/39
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\\ =\dfrac{1}{3}-\dfrac{1}{13}\\ =\dfrac{13}{39}-\dfrac{3}{39}\\ =\dfrac{10}{39}\)
NHỚ TÍNH ĐÚNG VỚI CHO COIN NHA
cách giải bài : 1^15+1^35+1^63+1^99+...+1^399+1^483. tính b
1/3 - 2/5 +3/35 - 4/63 + 5/99 - 6/143
2/35+2/63+2/99+2/143 = ?
2/3+2/15+2/35+2/63+2/99=
2/3 + 2/15 + 2/35 + 2/63 + 2/99
= 2/1×3 + 2/3×5 + 2/5×7 + 2/7×9 + 2/9×11
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11
= 1 - 1/11
= 10/11
2/3+2/15+2/35+2/63+2/99
=4/5+2/15+2/63+2/99
=14/15+2/63+2/99
=304/315+2/99
=1138/1155
2/3+2/15+2/35+2/63+2/99
\(=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{20}{11}\)
2/3+2/15+2/35+2/63+2/99
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
= \(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)
= \(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}\)
= \(\frac{3}{1\times3}-\frac{1}{1\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+...+\frac{11}{9\times11}-\frac{9}{9\times11}\)
= \(1-\frac{1}{11}\)=\(\frac{10}{11}\)