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đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}\)
\(=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)\)
\(100+\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}\right)-99=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1\)
\(=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)=100+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-99\)
\(=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100B\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
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\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)
\(\Rightarrow x=1\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
a) Đặt B = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)
\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right)\)
Đặt C = \(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\)
\(=\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)\)
\(=2\cdot\frac{1}{1.99}+2\cdot\frac{1}{3.97}+...+2\cdot\frac{1}{49.51}\)
\(=2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)
Thay B và C vào A
\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)
b) Đặt E = \(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)
\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Thay E vào B
\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\) Nhân với 2 cả hai vế:
được: \(2M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\) Suy ra : \(M=2M-M=1-\frac{1}{2^{100}}\)
CHÚC BẠN HỌC GIỎI
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
\(P=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)
\(2P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2P-P=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(P=\frac{1}{2}-\frac{1}{2^{100}}\)