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