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)308 exam-style questions on OCR A Level Maths 1.4 Sequences and Series, covering 1.4.1 Binomial expansion for positive integer n, 1.4.2 Link to binomial probabilities, 1.4.3 Binomial expansion for rational n (A-level only), 1.4.4 Validity of the expansion (A-level only), 1.4.5 Sequences (A-level only), 1.4.6 Increasing, decreasing and periodic sequences (A-level only), 1.4.7 Sigma notation (A-level only), 1.4.8 Arithmetic sequences and series (A-level only), 1.4.9 Geometric sequences and series (A-level only), 1.4.10 Sum to infinity of a geometric series (A-level only), and 1.4.11 Modelling with sequences and series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.