\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(\frac{1}{3.5.}\right).....\left(1+\frac{1}{99.101}\right)\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(=\frac{2^2.3^2.4^2.5^2.....98^2.99^2.100^2}{1.2.3^2.4^2.5^2......99^2.100.101}\)

\(=\frac{2.100}{1.101}\)

\(=\frac{200}{101}\)

1-1/101=100/101

nha bạn         

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....+\left(1+\frac{1}{99.101}\right)\)

\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\)

\(=100.\frac{2}{101}=\frac{200}{101}\)

Xét số hạng tổng quát: 
1 + 1/[k.(k + 2)] = [k.(k + 2) + 1]/[k.(k + 2)] = (k + 1)²/[k.(k + 1)], với k nguyên dương. 
Cho k chạy từ 1 đến 99, ta có: 
• 1 + 1/1.3 = 2²/(1.3). 
• 1 + 1/2.4 = 3²/(2.4). 
• 1 + 1/3.5 = 4²/(3.5). 
....................... 
• 1 + 1/97.99 = 98²/(97.99). 
• 1 + 1/98.100 = 99²/(98.100). 
• 1 + 1/99.101 = 100²/(99.101). 
Nhân vế với vế các đẳng thức trên, ta được: 
(1 + 1/1.3).(1 + 1/2.4)(1 + 1/3.5)....(1 + 1/99.101) 
= [2².3².....100²]/[1.2.3².4²......99².100...‡ 
= (2².100²)/(2.100.101) 
= 2.100/101 
= 200/101.

còn N thì chịu

M=(4/1.3.9/2.4.16/3.5...10000/99.101

M=2.2/1.3.3.3/2.4.4.4/3.5...100.100/99.101

M=2.3.4.5...100/1.2.3...99.3.4.5...100/2.3.4.5...101

M=100.2/101=200/101

Cau N sai de rui ban a, o mau so phai la 1.5.7+2.10.14+4.20.28+7.35.49 moi lam dc.

mấy bn giải khó hiểu wa .....................

= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101

= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101

= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)

= 100.2/101

=200/101

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)

\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)

\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)

\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)

\(A=\left(1+\frac{1}{1\cdot3}\right)\)\(\left(1+\frac{1}{2\cdot4}\right)\)\(\left(1+\frac{1}{3\cdot5}\right)\)\(......\left(1+\frac{1}{99\cdot101}\right)\)

\(=\frac{4}{1\cdot3}\)\(\cdot\frac{9}{2\cdot4}\)\(\cdot\frac{16}{3\cdot5}\)\(\cdot\cdot\cdot\cdot\cdot\frac{10000}{99\cdot101}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(=\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99\cdot101}\cdot\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{3\cdot4\cdot5\cdot\cdot\cdot\cdot99}\)

\(=\frac{1}{101}\cdot200\)

\(=\frac{200}{101}\)