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