A geometric progression has first term u1 u_1\,u1 and common ratio kkk, where k≠1k \ne 1k=1.
Show that the sum of the first n n\,n terms of this progression, SnS_nSn, can be expressed as
Sn=u1(1−kn)1−k S_n = \frac{u_1(1 - k^n)}{1 - k} Sn=1−ku1(1−kn)38 exam-style questions on AQA A Level Maths 1.7.5 Geometric sequences and series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.