Question

For a sequence, if S_n = 2 (3ⁿ – 1), find the n^th term, hence show that the sequence is a G.P.

Answer 1

S_n = 2(3ⁿ – 1)
∴ S_n–1 = 2(3ⁿ – 1 – 1)
But t_n = S_n – S_n–1 
= 2(3ⁿ – 1) – 2(3^n–1 – 1)
= 2(3ⁿ – 1 – 3^n–1 + 1)
= 2(3ⁿ – 3^n–1)
= 2(3^n–1+1 – 3^n–1)
∴ t_n = 2.3^n–1 (3 – 1) = 4.3^n–1
∴ t_n+1 = `4.3^(("n" + 1 - 1)`
= 4(3)ⁿ 
The sequence (t_n) is a G.P. if `("t"_("n" + 1))/"t"_"n"` = constant for all n ∈ N.

∴ `("t"_("n" + 1))/"t"_"n" = (4(3)^"n")/(4(3)^("n" - 1)` 

= 3 = constant, for all n ∈ N
∴ the sequence is a G.P. with t_n = 4(3)^n–1.