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Đặt 1.4+2.5+.....+n.(n+3)=C
Hướng dẫn: 3C = 3.[1.2 +2.3 +3.4 + ... + n(n + 1)] + 3.(2 + 4 + 6 + ... + 2n)
= 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n + 1).3 + 3.(2 + 4 + 6 + ... + 2n)
Nên C = n(n+1)(n+5):3
\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6+4.12.20+9.27.45}\)
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Ta thấy: 1.4 = 1.(1 + 3)
2.5 = 2.(2 + 3)
3.6 = 3.(3 + 3)
4.7 = 4.(4 + 3)
…….
n(n + 3) = n(n + 1) + 2n
Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n
C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n
C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)
⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n)
3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)
3C = n(n + 1)(n + 2) +
⇒ C = +
=
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Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3
2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4
...
Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)
N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)
Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
P=\(\dfrac{1}{2}-\dfrac{1}{100}\)
Vậy N=98-1+\(\dfrac{1}{50}\)
N=\(97+\dfrac{1}{50}\)
Vậy 97<N<98(ĐPCM)
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1.
`16 + (27 - 7.6 ) - (94 -7 - 27.99)`
`= 16+ 27 - 7.6 - 94 + 7 + 27.99`
`= 16 + 27(99 +1) - 7(6-1) - 94`
`= -78 + 27.100 - 7.5`
`= 2587`
2.
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`
`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`
`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`
`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`3/2A = 1 - 1/100`
`3/2 A= 99/100`
`A= 99/100 : 3/2`
`A=33/50`
Vậy `A= 33/50`
1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99
=(27+27.99)+(27+7-94)+16
=27.100-60+16
=2700-44=2656
2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(2.5+1.4-1\dfrac{3}{7}=\dfrac{5}{2}+\dfrac{7}{5}-\dfrac{10}{7}=\dfrac{5\cdot35}{2\cdot35}+\dfrac{7\cdot14}{5\cdot14}-\dfrac{10\cdot10}{7\cdot10}=\dfrac{175}{70}+\dfrac{98}{70}-\dfrac{100}{70}=\dfrac{173}{70}\)
2,5 + 1,4 - \(1\dfrac{3}{7}\)
=3,9 + \(\dfrac{10}{7}\)
=\(\dfrac{373}{70}\)