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(3^n–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–2

The sequence (t_n) is a G. P.,

If `"t"_"n"/"t"_("n"-1)` = constant

for all n ∈ N

∴ `"t"_"n"/"t"_("n" - 1) = (4.3^("n" - 1))/(4.3^("n" - 2))`

= `3^("n" - 1)/(3^("n" - 1)*3^((-1))`

= 3

= constant for all n ∈ N

∴ r = 3

∴ the sequence is a G.P. with t_n = 4.3^n–1