Question

Find S_n of the following arithmetico - geometric sequence: 

2, 4x, 6x², 8x³, 10x⁴, …

Answer 1

The numbers 2, 4, 6, 8, 10, ... form an A.P. whose first term is a = 2 and the common difference is d = 2.

Hence, the n^th term of this AP. is

a + (n – 1)d = 2 + (n – 1)2 = 2n

The numbers 1, x, x², x³, x⁴, ... form the G.P. whose first term is A = 1 and common ratio is r = x.

Hence, the n^th term of this G.P. is r^n–1 = x^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

t_n =[a + (n – 1)d]r^n–1 = (2n)x^n–1

The sum of first n terms of the series is

S_n = 2 + 4x + 6x² + 8x³ + 10x⁴ + ... +2(n – 1)x^n–2 +(2n)x^n–1   ...(1)

∴ x·S_n = 2x + 4x² + 6x³ + 8x⁴ + 10x⁵ + ... + 2(n – 1)x^n–1 + (2n)xⁿ   ...(2)

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

S_n – x·S_n = 2 + 2x + 2x² + 2x³ + ... + 2x^n–1 – (2n)xⁿ

∴ (1– x)S_n = 2 + 2x(1 + x + x² + ... + x^n–2) – (2n)xⁿ

= `2 + 2x[(1 - x^("n" - 1))/(1 - x)] - (2"n")x^"n"`

= `(2 - 2"n"x^"n") + (2x(1 - x^("n" - 1)))/(1 - x)`

∴ S_n = `(2(1 - "n"x^"n"))/(1 - x) + (2(1 - x^("n" - 1)))/(1 - x)^2`

Alternative Method:

The given sequence is

2, 4x, 6x², 8x³, 10x⁴, ...

This is arithmetico-geometric sequence with

a = 2, d = 2, r = x

S_n of this AG.P. is given by

S_n = `"a"/(1 - "r") + ("dr"(1 - "r"^("n" - 1)))/(1 - "r")^2 - (["a" + ("n" - 1)"d"]"r"^"n")/(1 - "r")`

= `2/(1 - x) + (2x(1 - x^("n" - 1)))/(1 - x)^2 - ([2 + ("n" - 1)2]x^"n")/(1 - x)`

= `2/(1 - x) + (2x(1 - x^("n" - 1)))/(1 - x)^2 - (2"n"*x^"n")/(1 - x)`

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