The sum of series 1.2.4 + 2.3.5 + 3.4.6 + ----------- 20 terms.
nth terms of the given series
tn = n(n+1)(n+3)
= n(n2 + 4n + 3)
= n3 + 4n2 + 3n
Sum to n terms sn = ∑ni=1 tn
Sn = ∑ni=1 (n3 + 4n2 + 3n) = ∑ni=1n3 + 4∑ni=1n2 + 3∑ni=1n
= n2(n+1)24+4.(n)(n+1)(2n+1)6+3n(n+1)2
Sn =n(n+1)12 [3n(n + 1) + 8(2n + 1) + 18]
Sn =n(n+1)12 [3n2 + 19n + 26]
= n(n+1)(n+2)(3n+13)12 --------------------(1)
Substituting n = 20, in the equation 1
S20 = 20(21)(22)(73)12 = 56210