Question

Find S_n of the following arithmetico - geometric sequence:

3, 12, 36, 96, 240, …

Answer 1

3, 12, 36, 96, 240, …

i.e., 1 × 3, 2 × 6, 3 × 12, 4 × 24, 5 × 48, …

Here, 1, 2, 3, 4, 5, … are in A.P.

∴ n^th term = n

3, 6, 12, 24, 48, … are in G.P.

∴ a = 3, r = 2

∴ n^th term = ar^n–1 = 3.(2^n–1)

∴ n^th term of the arithmetico-geometric sequence is t_n = n.3(2^n–1)

∴ S_n = 1 × 3 + 2 × 6 + 3 × 12 + 4 × 24 + 5 × 48 + ... + n.3(2^n–1)   ...(i)

Multiplying throughout by 2, we get

2S_n = 1 × 6 + 2 × 12 + 3 × 24 + 4 × 48 + 5 × 96 + …+ n.3(2ⁿ)   ...(ii)

Equation (i) – equation (ii), we get

S_n – 2S_n = [3 – 3n(2ⁿ)] + [6 + 12 + 24 + ... + 3(2^n–1)]

∴ – Sn = `[3 - 3"n"(2^"n")] + (6(2^("n" - 1) - 1))/(2 - 1)`

= 3 –  3n(2ⁿ) + 6(2^{n– 1} –  1)

∴ S_n = 3n(2ⁿ) – 3 – 6(2^n–1) + 6

= 3n(2ⁿ) – 3(2ⁿ) + 3

∴ S_n = (n – 1)3(2ⁿ) + 3 = 3[(n – 1)2ⁿ + 1]