Cho dãy số : \(\frac{1}{1\cdot3};\frac{1}{5\cdot7};\frac{1}{9\cdot11};\frac{1}{13\cdot15};...;\frac{1}{101\cdot103}\)
Số số hạng của dãy số trên là
Tính :
1. \(F=\frac{7}{90\cdot94}+\frac{7}{94\cdot98}+...+\frac{7}{158\cdot162}\)
2. \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{\text{4}\cdot5\cdot6}+...+\frac{1}{37\cdot38\cdot39}\)
F=7/4(1/90-1/94+1/94-1/98+...+1/158-1/162)
=7/4(1/90-1/162)
=7/4.2/405
=7/810
Vậy F=7/810
đặt A=1/1.2.3+1/2.3.4+...+1/37.38.39
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/37.38-1/38.39
2A=1/1.2-1/38.39
2A=1/2-1/1482
2A=370/741
A=370/741.1/2
A=185/741
HOÀN THÀNH
Tìm số tự nhiên x biết:
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
chứng tỏ \(\frac{1}{1\cdot2}+\frac{1}{1\cdot2\cdot3}+\frac{1}{1\cdot2\cdot3\cdot4}+...+\frac{1}{1\cdot2\cdot3....100}< 1\)
chứng tỏ \(\frac{1}{1\cdot2}+\frac{1}{1\cdot2\cdot3}+\frac{1}{1\cdot2\cdot3\cdot4}+...+\frac{1}{1\cdot2\cdot3\cdot...\cdot100}< 1\)
Tìm số tự nhiên x, biết: \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
Gọi A=(1/1.2.3+ 1/2.3.4 +...+ 1/8.9.10) .x=23/45
2A=3-1/1.2.3+ 4–2/2.3.4+ 5–4/3.4.5+ ... + 10–8/8.9.10
2A=1/2 —1/2.3+ 1/2.3 — 1/3.4+ 1/3.4– 1/4.5 +...+1/8.9–1/9.10=1/2–1/9.10=44/90
A=44/90 : 2=22/90
x=23/45:A= 23/45 : 22/90=23/11= 2 1/1( hỗn số)
1) Tìm n thuộc z để : \(\frac{19}{n-1}\cdot\frac{n}{9}\)có giá trị là số nguyên
2) Cho dãy:\(\frac{2}{11\cdot16};\frac{2}{16\cdot21};\frac{2}{21\cdot26};.....\).Tính tổng 50 phân số đầu tiên của dãy.
3) So sánh :
A = \(\frac{1\cdot3\cdot5\cdot.....\cdot2013}{1008\cdot1009\cdot1010\cdot....\cdot2014}\)và B = \(\frac{1}{2^{1007}}\)- 1
4) CMR: \(\frac{3}{1^2+2^2}+\frac{3}{2^2+3^2}+.....+\frac{19}{9^2+10^2}\)< 1
Các bạn ơi giúp mình với. Mình đang cần gấp lắm.
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Số k trong đẳng thức trên có giá trị là ?
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
TÍnh Tổng sau : \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot3}+...+\frac{1}{a+\left(a+1\right)+\left(a+2\right)}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(=\frac{n.\left(n+3\right)}{4.\left(n+1\right).\left(n+2\right)}\)
Bạn ghi rõ cách làm cho mình đc ko minh ko hiểu
ok ok , đề bài sai : Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{a.\left(a+1\right).\left(a+2\right)}\)
\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{a.\left(a+1\right)}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(A=\frac{1}{1.2}-\frac{1}{\left(a+1\right).\left(a+2\right)}=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Vậy \(A=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Cơ bản thì là vậy , nhưng cái phần tách ra mình ko nhớ , mình nhớ ra sẽ thông báo sao nhé