Sum to n terms of the series 11⋅2⋅3⋅4+12⋅3⋅4⋅5+13⋅4⋅5⋅6+⋯is

Let tr denote the rth term of the given series, then                tr=1r(r+1)(r+2)(r+3) and             tr+1=1(r+1)(r+2)(r+3)(r+4)⇒  rtr=1(r+1)(r+2)(r+3) and            (r+4)tr+1=1(r+1)(r+2)(r+3)Thus,    rtr−(r+4)tr+1=0⇒     rtr−(r+1)tr+1=3tr+1Putting    r=1,2,…,n−2,n−1, we get               1t1−2t2=3t22t2−3t3=3t33t3−4t4=3t4⋯(n−2)tn−2−(n−1)tn−1=3tn−1(n−1)tn−1−ntn=3tnAdding above expressions, we get                t1−ntn=3t2+t3+…+tn⇒         4t1−ntn=3t1+t2+…+tn⇒        11⋅2⋅3−1(n+1)(n+2)(n+3)=3Sn⇒       Sn=1316−1(n+1)(n+2)(n+3)