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)326 exam-style questions on Edexcel A Level Maths Sequences and Series, covering 3.1 Arithmetic Sequences, 3.2 Arithmetic Series, 3.3 Geometric Sequences, 3.4 Geometric Series, 3.5 Sum to Infinity, 3.6 Sigma Notation, 3.7 Recurrence Relations, and 3.8 Modelling with Series. Each one has a worked solution and a mark scheme showing where the marks go.