Question Number 10485 by Saham last updated on 13/Feb/17

A sequence of numbers T_1 , T_2 , T_3 , ...  Satisfies the  relation 3T_n = T_(n − 1)  + 6, Show that for all values   of n ≥ 1, T_n − 3  = (1/3)(T_(n − 3)  − 3). Hence or otherwise  show that if T_1  = 4, T_n  = 3 + 3^(1 − n)   for all values of  n, also find the sum of the first five terms of the   sequence.

Answered by mrW1 last updated on 19/Feb/17

3T_n = T_(n − 1)  + 6 ⇒T_(n−1) =3T_n −6    ...(i)  3T_(n−1) = T_(n −2)  + 6    ...(ii)  3T_(n−2) = T_(n − 3)  + 6 ⇒T_(n−2) =(T_(n−3) /3)+2     ...(ii)  (i) and (iii) in (ii):  3(3T_n −6)=((T_(n−3) /3)+2)+6  9T_n =(T_(n−3) /3)+26    T_1 =4=3+1=3+3^0   T_2 =(1/3)T_1 +2=(4/3)+2=3+3^(−1)   T_3 =(1/3)T_2 +2=(1/3)(3+3^(−1) )+2=3+3^(−2)   T_4 =(1/3)T_3 +2=(1/3)(3+3^(−2) )+2=3+3^(−3)   ...  T_n =3+3^(1−n)   S(n)=Σ_(k=1) ^n T_k =Σ_(k=1) ^n (3+3^(1−k) )=3n+1×((1−((1/3))^n )/(1−(1/3)))  =3n+((3^n −1)/3^n )×(3/2)  =3n+((3^n −1)/(2×3^(n−1) ))  S(5)=3×5+((3^5 −1)/(2×3^(5−1) ))=15+((243−1)/(2×81))=15+((121)/(81))=16((40)/(81))≈16.494

Commented bySaham last updated on 19/Feb/17

God bless you sir. i will reduce it.