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\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(=\frac{1}{100}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)
Đặt : \(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)
\(A=\frac{1}{100}\)
Vậy : \(A=\frac{1}{100}\)
D = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1.\right)\)
=>\(-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{100^2}.\right)\)
=>\(-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
=>\(-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}\right)\)
=>\(-\left(\frac{1.2.3...99}{2.3.4....100}\right)\left(\frac{3.4.5....101}{2.3.4....100}\right)\)
=>\(-\left(\frac{1}{100}.\frac{101}{2}\right)\)
=>\(D=-\frac{101}{200}\)
\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)
\(Q=\frac{1}{100}\)
\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)
\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)
\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)
Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới
\(P=\frac{201}{100}\)
Câu 1 Tính
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)
Câu 2 Tính
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)
\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
Câu 3
a) Ta có : M = 1 + 3 + 32 + 33 + ... + 3118 + 3119 (1)
=> 3M = 3 + 32 + 33 + 34 + ... + 3119 + 3120 (2)
Lấy (2) trừ (1) theo vế ta có :
3M - M = (3 + 32 + 33 + 34 + ... + 3119 + 3120) - ( M = 1 + 3 + 32 + 33 + ... + 3118 + 3119)
=> 2M = 3120 - 1
=> M = \(\frac{3^{120}-1}{2}\)
b) M = 1 + 3 + 32 + 33 + ... + 3118 + 3119
= (1 + 3 + 32) + (33 + 34 + 35) + ... + (3117 + 3118 + 3119)
= (1 + 3 + 32) + 33(1 + 3 + 32) + ... + 3117(1 + 3 + 32)
= 13 + 33.13 + ... + 3117.13
= 13(1 + 33 + ... + 3117) \(⋮\)13
=> M \(⋮\)13
M = 1 + 3 + 32 + 33 + ... + 3118 + 3119
= (1 + 3 + 32 + 33) + (34 + 35 + 36 + 37) + ... + (3116 + 3117 + 3118 + 3119)
= (1 + 3 + 32 + 33) + 34(1 + 3 + 32 + 33) + ... + 3116(1 + 3 + 32 + 33)
= 40 + 34.40 + ... + 3116.40
= 40(1 + 34 + ... + 3116)
= 5.8.(1 + 34 + ... + 3116) \(⋮\)5
4) Tính
A = 2100 - 299 - 298 - ... - 22 - 2 - 1
=> 2A = 2101 - 2100 - 299 - 298 - 22 - 2 - 1
Lấy 2A trừ A theo vế ta có :
2A - A = (2101 - 2100 - 299 - 298 - 22 - 2 - 1) - (2100 - 299 - 298 - ... - 22 - 2 - 1)
=> A = 2101 - 2100 - 2100 + 1
=> A = 2101 - (2100 + 2100) + 1
=> A = 2101 - 2100 . 2 + 1
=> A = 1
Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
= 99.100.101
=> C = 99.100.101 : 3 = 333300
b) Ta có : D = 22 + 42 + 62 + ... + 982
= 22(12 + 22 + 32 + ... + 492
= 22 .(12 + 22 + 32 + ... + 492)
= 22.(1.1 + 2.2 + 3.3 + ... + 49.49)
= 22.[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]
= 22.[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]
Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50
= 49.50.51
=> E = 49.50.51/3 = 41650
Khi đó D = 22.[41650 - (1 + 2 + 3 + 4 + ... + 49)]
= 22.[41650 - 49(49 + 1)/2]
= 22.[41650 - 1225
= 22.40425
= 161700
=> D = 161700
\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)
\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)
\(=0\)
Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{n^2-1}{n^2}\)
\(\Rightarrow A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
\(\Rightarrow A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\cdot\cdot\left[\left(n-1\right)\left(n+1\right)\right]}{\left(2\cdot2\right)\cdot\left(3\cdot3\right)\cdot\cdot\cdot\left(n\cdot n\right)}\)
\(\Rightarrow A=\frac{\left[1\cdot2\cdot\cdot\cdot\cdot\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot\cdot\cdot\cdot\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)}\)
\(\Rightarrow A=\frac{1\cdot\left(n+1\right)}{n\cdot2}\)
\(\Rightarrow A=\frac{n+1}{2n}\)
A=(1-1/2^2)(1-1/3^2).....(1-1/n^2)
A=1(1/2^2-1/3^2-...-1/n^2)
......
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