Mathematics

Mathematics

Problem

Let aa be the first term and r (≠1)r\,(\neq1) the common ratio of a geometric progression. The sum of the first nn terms is denoted by SnS_n. a. (i) Show that S2n=Sn+r n SnS_{2n}=S_n+r^{\,n}\,S_n. b. (ii) Hence obtain S2nSn\displaystyle\frac{S_{2n}}{S_n} in terms of rr and nn. c. (iii) Taking r=2r=2, determine the least positive integer nn for which S2nS_{2n} exceeds SnS_n by more than 999a999a. Show all steps clearly.
Intuitive explanation

The key is recognizing that a geometric series has a closed-form formula that collapses an infinite-looking sum into something manageable. When you multiply SnS_n by the common ratio rr and subtract, most terms cancel—this telescoping trick reveals the pattern. Understanding why this works (each term shifts position) unlocks not just the formula itself, but intuition for handling any geometric series problem.

Intuitive explanation

The key is recognizing that a geometric series has a closed-form formula that collapses an infinite-looking sum into something manageable. When you multiply SnS_n by the common ratio rr and subtract, most terms cancel—this telescoping trick reveals the pattern. Understanding why this works (each term shifts position) unlocks not just the formula itself, but intuition for handling any geometric series problem.

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