Bài 1. Tính `B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)`
Bài 2. Tính `C = 1.4 + 2.5 + 3.6 + 4.7 + … + n(n + 3)`
Bài 1. Tính `B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)` Bài 2. Tính `C = 1.4 + 2.5 + 3.6 + 4.7 + … + n(n + 3)`
By Arya
By Arya
Bài 1. Tính `B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)`
Bài 2. Tính `C = 1.4 + 2.5 + 3.6 + 4.7 + … + n(n + 3)`
Đáp án+Giải thích các bước giải:
`B=1.2.3 + 2.3.4 + . .. + (n – 1)n(n + 1)`$\\$`=>4B=1.2.3.4 + 2.3.4.(5-1) + . .. + (n – 1)n(n + 1).[(n+2)-(n-2)`$\\$`=>4B=1.2.3.4+2.3.4.5-1.2.3.4+.. .+ (n – 1)n(n + 1).(n+2)-(n-2).(n – 1)n(n + 1).`$\\$`=>4B=(n – 1)n(n + 1).(n+2)`$\\$`=>B=((n – 1)n(n + 1).(n+2))/4`
`C = 1.4 + 2.5 + 3.6 + 4.7 + … + n(n + 3)`$\\$`C=1.2+2+2.3+4+3.4+6+. . .+n ( n + 1 ) + 2 n`$\\$`C=[1.2+2.3+3.4+ .. .+n(n+1)]+(2+4+6 + . .. +2n)`$\\$`Đặt,3B={1.2.3+2.3.(4-1)+3.4.(5-2)+ .. .+n(n+1).[(n+2)-(n-1)]}`$\\$`=>B=(n.(n+1).(n+2))/3`$\\$`=>C=(n.(n+1).(n+2))/3+n.(n+1)`$\\$`=>C=n.(n+1).[(n+2)/3+1]`
Bài 1
`4B = 1.2.3.4 + 2.3.4.4 + … + (n – 1)n(n + 1).4`
`4B = 1.2.3.4 – 0.1.2.3 + 2.3.4.5 – 1.2.3.4 + … + (n – 1)n(n + 1)(n + 2) – [(n – 2)(n – 1)n(n + 1)]`
`4B = (n – 1)n(n + 1)(n + 2) – 0.1.2.3 = (n – 1)n(n + 1)(n + 2)`
=> B = $\frac{( n-1)n(n-1)(n-2)}{4}$
Bài 2
Ta thấy`: 1.4 = 1.(1 + 3)`
`2.5 = 2.(2 + 3)`
`3.6 = 3.(3 + 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)` +$\frac{3(n+2)n}{2}$
C=$\frac{n(n+1)(n+5)}{3}$