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Question

If Sn and Tn represent the sum of n terms and the nth term respectively, of the series 1+4+10+20+35+, then the value of 19T20S19 is

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Solution

Sn=1+4+10+20+35++TnSn= 1+4+10+20+35++Tn1+Tn

Tn=1+3+6+10+15++tn, where tn=TnTn1
Tn= 1+3+6+10+15++tn1+tn

tn=1+2+3+4+upto n termstn=12n(n+1)Tn=nr=0tr=12nr=0r(r+1)Tn=12[n(n+1)(n+2)3]Tn=16n(n+1)(n+2)

Sn=nr=0TrSn=16nr=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

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