1+1000-1000
1/1000+13/1000+25/1000+37/1000+49/1000+...+87/1000+99/1000
1/1000+13/1000+25/1000+37/1000+49/1000+......+87/1000+99/1000
Ta có : 1+13+25+37+...+99/1000
Ta tính tử số
Số số hạng của tử số là : (99-1)/12+1
Sai đề bạn ơi
1/1000 + 13/1000+25/1000+37/1000+49/1000+...+87/1000+99/1000
khanh rat de doc ki de roi neu ko hieu thi nhan tin hoi minh
Viết đầy đủ các số hạng ta có:
1/1000 + 13/1000 + 25/1000 + 37/1000 + 49/1000 + 51/1000 + 63/1000 + 75/1000 + 87/1000 + 99/1000
= 1 + 13 + 25 + 37 + 49 + 51 + 63 + 75 + 87 + 99/ 1000
= ( 1 + 99 ) + ( 13 + 87) + ( 25 + 75 ) + ( 37 + 63) + ( ( 49 + 51)/ 1000
= 5 x 100/ 1000
= 500/ 1000
= 1/2
\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+\frac{49}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)
1/1000 + 13/1000 + 25/1000 + 37/1000 + +49/1000 + ............. + 97/1000 + 109/1000
\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+....+\frac{109}{1000}\)
\(=\frac{1+13+25+....+109}{1000}\)
Áp dụng công thức tính dãy số ta có
\(1+13+25+...+109=\frac{\left[\left(109-1\right):12+1\right].\left(109+1\right)}{2}=\frac{10.110}{2}=10.55=550\)
Vậy
\(\frac{1+13+25+...+109}{1000}=\frac{550}{1000}=\frac{11}{20}\)
\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+.......+\frac{109}{1000}\)
\(\frac{1+13+25+37+.....+97+109}{1000}\)
\(\frac{\left(\left(109-1\right):12+1\right).\left(109+1\right):2}{1000}\)
\(\frac{550}{1000}\)
= \(\frac{11}{20}\)
Cho A=1001/1000*1000+1 + 1001/1000*1000+2 + ...... + 1001/1000*1000+1000
Chứng minh: 1<A*A<4
1 / 1000 + 13 / 1000 + 25 / 1000 + 49 / 1000 + ... + 87 / 1000 + 99 / 1000
Bằng 2,5 = \(\frac{2500}{1000}\)= \(\frac{25}{10}\) nhé
1/1000+13/1000+25/1000+37/1000+49/1000+....99/1000=?
Tính nhanh : M =1/1000+13/1000+25/1000+37/1000+...+121/1000+133/1000
M =1/1000+13/1000+25/1000+37/1000+...+121/1000+133/1000
\(M=\frac{1+13+25+...+133}{1000}\)
\(M=\frac{\left(133+1\right)\times12:2}{1000}\)
\(M=\frac{804}{1000}=0,804\)
Tinh [(1+2012/1)*(1+2012/2)*(1+2012/3)*...*(1+2012/1000)]/[(1+1000/1)*(1+1000/2)*(1+1000/3)*...*(1+1000/2012)]
C ={(1+(1999/1))(1+(1999/2))(1+(1999/3))+...+(1+(1999/1000))}/{(1+(1000/1))(1+(1000/2))(1+(1000/3))...(1+(1000/1999))}