Sum to n terms of the series 1+(1+2)+(1+2+ 3) +…is
12n(n+1)(n+2)
13n(n+1)(n+2)
14n(n+1)(2n+1)
16n(n+1)(n+2)
tr=1+2+…+r=12r(r+1)=12r2+r∑r=1n tr=1216n(n+1)(2n+1)+12n(n+1)=16n(n+1)(n+2)