1×3 + 3×5 + 5×7 + ... + 99×101.
2/1×3+2/3×5+.......+2/99×101
5/1×3+5/3×5+5/5×7+.......+5/99×101
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)
\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)
\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
....
\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)
=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)
=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)
=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
=\(\frac{5}{2}\cdot\frac{100}{101}\)
\(=\frac{250}{101}\)
= 3 - 1 / 1 x 3 + 5 - 3 / 3 x 5 + ... + 101 - 99 / 99 x 101
= 1 - 1 / 3 + 1 / 3 - 1 / 5 + 1 / 5 - ... - 1 / 99 + 1 / 99 - 1 / 101
gạch gạch gạch gạch ... gạch gạch
= 1 - 1 / 101
= 100 / 101
Bài tập *
a) 2/1×3 + 2/3×5 + 2/5×7 + ...............+ 2/99×101
b) 5/1×3 + 5/3×5 + 5/ 5×7 + ................ + 5/99×101
Giúp mk bài này nhé mk
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
a) A= 1+(-2)+(-3)+4+5(-6)+(-7)+8+9+...+99+100-101+102+103
b) B=1+(-3)+5+(-7)+...+57+(-99)+101
1*3+3*5+5*7+...+97*99+99*101
A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)
A = (12 + 32 + 52 + … + 972 + 992) + 2.(1 + 3 + 5 + … + 97 + 99).
Đặt B = 12 + 32 + 52 + … + 992
=> B = (12 + 22 + 32 + 42 + … + 1002) – 22.(12 + 22 + 32 + 42 + … + 502)
Tính dãy tổng quát C = 12 + 22 + 32 + … + n2
C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]
C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)
C = = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6
Áp dụng vào B ta được:
B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650
=> A = 166650 + 2.(1 + 99).50 : 2
=> A = 166650 + 5000 = 172650.
Vậy: A = 172650.
Chúc bạn học tốt !!!
Tính:
a, A= 1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+99-100-101+102+103
b,B=1+(-3)+5+(-7)+...+97+(-99)+101
Tính:
1*3+3*5+5*7+...+97*99+99*101
Tính B=1*3+5*7+9*11+...+97*101
C=1*3*5-3*5*7+5*7*9-....-97*99*101
D=1*99+3*97+5*95+...+49*51
E=1*3^3+3*5^3+5*7^3+...+49*51^3
F=1*99^2+2*98^2+3*97^2+...+49*51^2
cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà
cho S = 1/1*3+1/3*5+1/5*7+...+1/99*101.Khi đó 2S + 1/101=
2S = \(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}\)
\(2S=1-\frac{1}{101}\)
2S + 1/101 = \(1-\frac{1}{101}+\frac{1}{101}=1\)
A = 1-2+3-4+5-6+.....+97-98+99-100+101
B = 1+2-3-4+5+6-7-8+......+98-99-100+101
B=1+2-(3+4)+5+6-..-100+101
B=(3+11+19+...+195)-(7+15+...+199)+101
B=25.99-25.103+101
B=-100+101=1
Vậy B=1
a=151
B=1 nha
5/1×3+5/3×5+5/5×7+...+5/99×101
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)