If the sum to first n terms of a series, the #r^(th)# term of which is given by #(2r + 1)^(2r)# can be expressed as #R(n*2^n) + S*2^n + T#, then find the value of (R+T+S)?
Given that the #r^(th)# term of the seriess #t_r=(2r + 1)^(2r)...[1]#
Inserting #r=1,2and 3# in [1] we get
#t_1=(2*1 + 1)^(2*1)=9#
#t_2=(2*2 + 1)^(2*2)=625#
and
#t_3=(2*3 + 1)^(2*3)=117649#
Again sum of first n terms
#S_n=R(n*2^n) + S*2^n + T#
So #t_1=S_1=2R+2S+T#
#=>2R+2S+T=9.....[2]#
#t_2=S_2-S_1#
#=>t_2=[R(2*2^2) + S*2^2+T]-[2R+2S+T]#
#=>t_2=6R+2S#
So #6R+2S=625....[3]#
#t_3=S_3-S_2#
#=>t_3=[R(3*2^3) + S*2^3+T]-[R(2*2^2) + S*2^2+T]#
#=>t_3=16R+4S#
#=>16R+4S=117649....[4]#
From [3] and [4] we get
#4R=117649-1250#
#=>R=29099.75....[5]#
From [5] and [3] we get
#6xx29099.75+2S=625#
#=>S=-86986.75#
Inserting the value of R and S in [2] we get
#=>2xx29099.75-2xx86986.75+T=9#
#T=115765#
Hence
#R+T+S=29099.75+115765-86986.75=57878#
