Sum to 30 terms of the series 11.2.3+12.3.4+13.4.5+… is

tr=1r(r+1)(r+2)=12r−1r+1+12(r+2)[split into partial fractions] =121r−1r+1−121r+1−1r+2⇒ Sn=∑r=1n tr=121−1n+1−1212−1n+2=12nn+1−n4(n+2) Thus, S30=123031−304(32)=30464−31(31)(32)=4951984