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= 1/2 -1/4 + 1/4 - 1/6 + 1/6 + 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100
= 1/2 - 1/100
= 49/10
ta có : đề bài
=1/2*(2/2*4+2/4*6+2/6*8+...+2/96*98+2/98*100)
=1/2*(1/2-1/4+1/4-1/6+1/6-1/8+...+1/96-1/98+1/98-1/100)
=1/2*(1/2-1/100)
=1/2*49/10
=49/200
Bạn bên trên nhầm 49/100 thành 49/10. Kết quả đúng nhưng bạn ko chú ý dòng thứ 2 từ dưới lên
C= \(\frac{49}{200}\)
D= \(\frac{33}{100}\)
Chúc bạn Hk tốt!!!!
C =1/2*4+1/4*6+1/6*8+...+1/98*100
2xC=2/2*4+2/4*6+2/6*8+...+2/98*100
2xC=1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100
2xC=1/2-1/100
2xC=49/100
C=49/100:2
C=49/200
Ý B làm tương tự nhưng nhưng cả 2 vế với 3
nha. ^_^ ^_^ ^_^
\(\frac{1}{2\cdot4}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)
\(=\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}+\frac{2}{98\cdot100}\right]\)
\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right]\)
\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{100}\right]=\frac{1}{2}\left[\frac{50}{100}-\frac{1}{100}\right]=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{49}{100}\)
\(=\frac{49}{200}\)
~Học tốt~
1, số số hạng là :
(100 - 1) + 1=100 (số)
Tổng là :
( 100 + 1 )x 100 : 2 = 5050
2, Số số hạng
(100 - 2 ) : 2 +1 = 45(số)
Tổng là :
( 100 +2) x 45 :2 =2295
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha