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\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)
\(=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{98^2}\right)\left(1-\frac{1}{99^2}\right)\)
\(=\frac{3}{2^2}.\frac{8}{3^2}......\frac{9603}{98^2}.\frac{9800}{99^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)
\(=\frac{1.2.4...97.98}{2.3....98.99}.\frac{3.4...99.100}{2.3....98.99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{50}{99}\)
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
\(\frac{\frac{1.2}{2}+\frac{2.3}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}=\frac{1}{2}\)
Ta gọi:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+97\cdot98}\) là A
\(=\frac{1+3+6+10+...+4753}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(\Rightarrow2A=\frac{2+6+12+20+...+9506}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(=\frac{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
=> 2A = 1
=> A = 1/2
Mik nghĩ bỏ 98.98 ở phần MS thì bài mới đúng, bn nên hỏi lại thầy bn
Vậy A = 1/2
Lời giải:
Xét tử số:
$1+(1+2)+(1+2+3)+...+(1+2+3+....98)$
$=\underbrace{(1+1+.....+1)}_{98}+\underbrace{(2+2+...+2)}_{97}+....+\underbrace{97+97}_{2}+98$
$=1.98+2.97+3.96+...+98.1$
Do đó: $B=\frac{1.98+2.97+3.96+..+98.1}{1.98+2.97+3.96+...+98.1}=1$