Question

Find S_n of the following arithmetico - geometric sequence:

1, 2 × 3, 3 × 9, 4 × 27, 5 × 81, …

Answer 1

The numbers 1, 2, 3, 4, ... form an A.P. whose first term is a = 1 and the common difference is d = 1.

Hence, the n^th term of this A.P. is

a + (n – 1)d = 1 + (n – 1)1 = n

The numbers 1, 3, 9, 27, 81, ... form the G.P. whose first term is A = 1 and common ratio is r = 3.

Hence, the nth term of this G.P. is

r^n–1 = 3^n–1

The terms of the given sequence are obtained by multiplying the corresponding terms of the above A.P. and G.P. Hence, the given series is an arithmetico-geometric series whose nth term is

[a + (n – 1)d]r^n–1 = n·3^n–1

The sum of first n terms of the series is

S_n = 1 + 2 × 3 + 3 × 9 + 4 × 27 + 5 × 81 + ...... + (n – 1)·3^n–2 + n·3^n–1    ...(1)

∴ 3·S_n = 1 × 3 + 2 × 9 + 3 × 27 + 4 × 81 + ... + (n – 1)·3^n–1 + n·3ⁿ   ...(2)

Subtracting (2) from (1), we get,

S_n – 3·S_n = 1 + 1 × 3 + 1 × 9 + 1 × 27 + 1 × 81 + ... + 3^n–1 – n·3ⁿ 

∴ – 2·S_n = 1+(3 + 3² + 3³ + 3⁴ + ... + 3^n–1) – n·3ⁿ 

= 1 + 3(1+3 + 3² + 3³ + ... + 3^n–2) – n·3ⁿ

= `1 + 3((1 - 3^("n" - 1))/(1 - 3)) - "n".3^"n"`

= `1 - 3/2(1 - 3^("n" - 1)) - "n".3^"n"`

∴ S_n = `-1/2 + 3/4(1 - 3^("n" - 1)) + ("n".3^"n")/2`

= `("n".3^"n" - 1)/2 + (3 - 3"n")/4`