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

`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=$\frac{n(n+1)(n+5)}{3}$