Mathematics

Mathematics

Problem

Which of the following correctly gives the sum SnS_n of the first nn terms of a geometric progression whose first term is aa and common ratio r  (r≠1)r\;(r\neq1)? a) arn−11−ra\dfrac{r^{n}-1}{1-r} b) ar n−1(r−1)ar^{\,n-1}(r-1) c) arn−1r−1a\dfrac{r^{n}-1}{r-1} d) arn−1−1r−1a\dfrac{r^{n-1}-1}{r-1}
Intuitive explanation

A geometric series grows (or shrinks) by multiplying each term by the same ratio rr, so the sum involves adding up a+ar+ar2+⋯+arn−1a + ar + ar^2 + \cdots + ar^{n-1}. The key insight is to use a clever algebraic trick: multiply the entire sum by rr, subtract it from the original sum, and watch most terms cancel—this telescoping leaves you with just the first and last terms, which you can solve for SnS_n directly.

Intuitive explanation

A geometric series grows (or shrinks) by multiplying each term by the same ratio rr, so the sum involves adding up a+ar+ar2+⋯+arn−1a + ar + ar^2 + \cdots + ar^{n-1}. The key insight is to use a clever algebraic trick: multiply the entire sum by rr, subtract it from the original sum, and watch most terms cancel—this telescoping leaves you with just the first and last terms, which you can solve for SnS_n directly.

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