Find the general term and sum to n terms of the sequence `1, 4/3, 7/9, 10/27, ......`

#### Solution

The given sequence as `1, 4/3, 7/9, 10/27, ......`

Consider the terms in the numerator 1, 4, 7, 10, ……

Which is an Arithmetic progression with first term a = 1

Common difference d = 4 – 1 = 3

a_{n} = a + (n – 1)d

= 1 + (n – 1)^{3}

= 1 + 3n – 3

= 3n – 2

The given sequence can be written as `1, (1 + 3)(1/3)`,

`(1 + 2 xx 3)(1/3)^2`

`(1 + 3 xx 3) (01/3)^3, .......`

Where `1, 1/3, 1/3^2, 1/3^3, ......` is a G. P with first term a = 1, common ratio r = `1/3`

∴ The given sequence is an Arithmetic – Geometric sequence.

∴ T_{n} = `["a" + ("n" - 1)"d"]"r"^("n" - 1)`

T_{n} = `[1 + ("n" - 1)^3] (1/3)^("n" - 1)`

T_{n} = `[ 1 + 3"n" - 3] 1/(3^("n" - 1))`

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

∴ The given sequence is `1, ((6 - 2)/3^1), ((9 - 2)/3^2), ((12 - 2)/3), ..........`

`1, (4/3^1), (7/3^2), (10/3^3), .........`

`1, (1 + 3) 1/3, (1 + 2) 1/3^2, (1 + 3) 1/3^2, .........`

Which is an arithmetic – geometric sequence.

∴ The sum of first n terms of the arithmetico – geometric sequence is given by

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

Here a = 1, d = 3, r = `1/3`

= `(1 - (1 + ("n" - 1)3) 1/3^"n")/(1 - 1/3) + 3 xx 1/3 [((1 - (1/3)^("n" - 1)))/(1 - 1/3)]`

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

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

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