Example 5 a) The \( n^{6} \) term of an exponential sequence (GP) is \( 2^{n} \). Calculate: i) the sum \( S_{4} \), of the first four terms of the sequence, ii) the sum of the sequence for large positive integral values of \( n \). b) The sum of the first \( n \) terms of a series is given by \( S_{n}=n(n+2) \). Find an expression for the \( n \)th term.

We are given two separate problems. Read carefully and proceed as follows. ────────────────────────────── Problem (a) We are told that the nth term of a geometric progression (exponential sequence) is   aₙ = 2ⁿ. Recall that in a geometric progression (GP) the nth term is usually given by   aₙ = a · r^(n–1). Here we see that 2ⁿ can be written as 2 · 2^(n–1). Thus the first term and common ratio are   a = 2  and  r = 2. (i) Sum of the first four terms, S₄: We have   a₁ = 2,   a₂ = 2² = 4,   a₃ = 2³ = 8,   a₄ = 2⁴ = 16. Therefore,   S₄ = 2 + 4 + 8 + 16 = 30. (ii) Sum of the sequence for large positive integral values of n: Since the common ratio r = 2 is greater than 1, the terms in the GP increase without bound. That is, as n increases the terms become larger and the partial sum grows indefinitely. Hence, the sum of the infinite sequence diverges (it does not approach a finite limit). ────────────────────────────── Problem (b) We are told that the sum of the first n terms of a series is given by   Sₙ = n(n + 2). To find an expression for the nth term, recall that the nth term aₙ can be found by subtracting the (n–1)th partial sum from the nth partial sum:   aₙ = Sₙ – Sₙ₋₁. Now, calculate Sₙ₋₁:   Sₙ₋₁ = (n – 1)[(n – 1) + 2] = (n – 1)(n + 1). Then,   aₙ = n(n + 2) – (n – 1)(n + 1). Expand each product:   n(n + 2) = n² + 2n,   (n – 1)(n + 1) = n² – 1      [using the difference of squares] Thus,   aₙ = (n² + 2n) – (n² – 1) = n² + 2n – n² + 1 = 2n + 1. ────────────────────────────── Final Answers a)  (i) S₄ = 30.  (ii) The sum diverges; that is, as n becomes very large, the sum increases without limit. b)  The nth term is aₙ = 2n + 1.