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)?

1 Answer
Dec 6, 2017

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