Sn=1+4+10+20+35+⋯+TnSn= 1+4+10+20+35+⋯+Tn−1+Tn
Tn=1+3+6+10+15+⋯+tn, where tn=Tn−Tn−1
Tn= 1+3+6+10+15+⋯+tn−1+tn
⇒tn=1+2+3+4+⋯upto n terms⇒tn=12n(n+1)⇒Tn=n∑r=0tr=12n∑r=0r(r+1)⇒Tn=12[n(n+1)(n+2)3]⇒Tn=16n(n+1)(n+2)
Sn=n∑r=0Tr⇒Sn=16n∑r=0r(r+1)(r+2)⇒Sn=124n(n+1)(n+2)(n+3)
So, nTn+1Sn=16n(n+1)[(n+1)+1][(n+1)+2]124n(n+1)(n+2)(n+3)
∴19T20S19=4